PHYSICS
JEE MAINS
Dropper
& ADVANCED
MODULE - 02
Newton’s Laws of Motion and Friction
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JEE - PHYSICS
Newton’s Laws of Motion
and Friction
FORCE
TYPES OF FORCES :
1.
Contact forces :
Whenever two bodies come in contact they exert forces
on each other, that are called contact forces.
(a) Normal reaction (N or R) : It is the component of
contact force normal to the surface. It measures how
strongly the surfaces in contact are pressed together.
(b) Frictional force (f ) : It is the component of contact
force parallel to the surface. It opposes the relative
motion (or attempted motion) of the two surfaces in
contact.
(c) Tension : The force exerted by the end of a taut
string, rope or chain is called the tension. The direction
of tension is always pulling in nature.
4
(d) Spring force : Every spring resists any attempt to
change its length, the more you change its length the
harder it resists. The force exerted by a spring is given
by F = –kx, where x is the change in length and k is
spring constant or stiffness constant (units N/m).
A force changes the state of rest or motion of a body.
When a force act on a body it start moving if it is at rest
or if it is in motion then to stop it, or to deflect it from its
initial path of motion. Force is defined as an interaction
between two bodies.
BASIC FORCES IN NATURE
(i) Gravitational force
(ii) Electromagnetic force
(iii) Nuclear force
(iv) Weak force
 Other well known forces like frictional force, elastic
force, viscous force, spring force, intermolecular
force are manifestations of electromagnetic forces.
 Electromagnetic force is about 1036 times stronger
than gravitational force while weak forces are about
10 25 times as strong as gravitational force.
FN > FE > FW > FG
FG : FW : FE : FN : : 1 : 1025 : 1036 : 1038
 Gravitational and electromagnetic forces are central
or conservative forces and obey inverse square
law.
 The gravitational and the electromagnetic forces
are long range forces having infinite range while
nuclear and weak forces are very short range forces.
 The nuclear force is the strongest force while the
gravitational force is the weakest force of nature.
 Nuclear force is a non-central force and varies
inversely with some higher power of distance.
 Beta-decay can be explained only on the basis of
weak forces.
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2.
Non contact forces :
These forces do not involve physical contact between
two points and are also called field forces
e.g. Gravitational force,electric force.
Note :
Weight : Weight of a body is the force with which
earth attracts it. It is also defined as force of gravity or
the gravitational force.
NEWTON'S FIRST LAW OF MOTION
OR LAW OF INERTIA
Every body continues in its state of rest or uniform
motion in a straight line unless compelled by an external
force to change that state.
Definition of force from Newton's first law of motion
"Force is that push or pull which changes or tends to
change the state of rest or of uniform motion in a straight
line".
Inertia : Inertia is the property of the body due to
which body oppose the change of state by itself . Inertia
of a body is measured by mass of the body.
Inertia  mass
Heavier the body, greater is the force required to change
its state and hence greater is inertia. The reverse is also
true. i.e. lighter body has less inertia.
TYPES OF INERTIA
Inertia of rest : It is the inability of a body to change
by itself, its state of rest.
 When we shake a branch of a mango tree, the
mangoes fall down.

When a bus or train starts suddenly the passengers
sitting inside tends to fall backwards.

The dust particles in a blanket fall off when it is
beaten with a stick.
Newton’s Laws of Motion and Friction
JEE - PHYSICS


the same line. The two forces F and F connected by
Newton's third law are called action-reaction pair. Any
one may be called 'action' and the other 'reaction'.
“Action and reaction acts on different bodies
hence they never cancel each other”.
Inertia of motion : It is the inability of a body to change
by itself its state of uniform motion.

When a bus or train stop suddenly, a passenger
sitting inside tends to fall forward.

A person jumping out of a speeding train may fall
forward.
Inertia of direction : It is the inability of a body to
change by itself its direction of motion.

When a car rounds a curve suddenly, the person
sitting inside is thrown outwards.

Rotating wheels of vehicle throw out mud,
mudguard over the wheels stop this mud.
NEWTON'S SECOND LAW OF MOTION
Rate of change of momentum of a body is directly
proportional to the external force applied on it and the
change in momentum takes place in the direction of
force

 dp
F
dt
or



dv
Fm
 ma
dt
  dm
Fv
dt

 dp
F
dt
if m = constant
(as in case of conveyor belt)
Special points :
1. This law gives magnitude, unit and dimension of
force.
2. There are two types of unit of force :
(a) Absolute unit
(b) Gravitational or Practical unit
In M.K.S. : Newton (N)
In M.K.S. : kg-wt or kg-f
In C.G.S. : Dyne (dyne)
In C.G.S. : gm-wt or gm-f
1 kg-wt = 9.8 N
1 gm-wt = 980 dyne
NEWTON'S THIRD LAW OF MOTION
It states that to every action, there is an equal (in
magnitude) and opposite (in direction) reaction.

If a body A exerts of force F on another body B, then

B exerts a force F on A, the two forces acting along
Newton’s Laws of Motion and Friction
FREE BODY DIAGRAM
A free body diagram is a diagram showing the chosen
body by itself, "free" of its surroundings, with vectors
drawn to show the magnitudes and directions of all
the forces applied to the body by the various other
bodies that interact with it.
Be careful to include all the forces acting on the body,
but the equally careful not to include any force that
the body exerts on any other body. In particular, the
two forces in an action-reaction pair must never appear
in the same free-body diagram because they never act
on the same body.
[Forces that a body exerts on itself are never included,
since these can't affect the body's motion.]
NOTE :
 The forces exerted on bodies of a given system
by bodies situated outside are called external
forces.
 The forces of interaction between bodies
composing a system are called internal forces.
 A single isolated force is physically impossible.
 Whenever one force acts on a body it gives rise to
another force called reaction.
 Total internal force in an isolated system is always
zero.
PROBLEM SOLVING STRATEGY :
Newton’s laws refer to a particle and relate the forces
acting on the particle to its mass and to its acceleration.
But before writing any equation from Newton’s law,
you should be careful about which particle you are
considering. The laws are applicable to an extended
body too which is nothing but collection of a large
number of particles.
Follow the steps given below in writing the equations :
Step 1 : Select the body
The first step is to decide the body on which the laws
of motion are to be applied. The body may be a single
particle, an extended body like a block, a combination
of two blocks-one kept over another or connected by a
string. The only condition is that all the parts of the
body or system must have the same acceleration.
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JEE - PHYSICS
Step 2 : Identify the forces
Once the system is decided, list down all the force
acting on the system due to all the objects in the
environment such as inclined planes, strings, springs
etc. However, any force applied by the system
shouldn’t be included in the list. You should also be
clear about the nature and direction of these forces.
Step 3 : Make a Free-body diagram (FBD)
Make a separate diagram representing the body by a
point and draw vectors representing the forces acting
on the body with this point as the common origin.
This is called a free-body diagram of the body.
(3)
The necessary condition for the equilibrium of a body
under the action of concurrent forces is that the vector sum of all the forces acting on the body must be
zero.
(4)
Mathematically for equilibrium
F
x
(5)
0 ;
F
y
z
R
wp
R
F.B.D. of platform
C
TC
wm
F.B.D. of man
Look at the adjoining free-body diagrams for the
platform and the man. Note that the force applied by
the man on the rope hasn’t been included in the FBD.
Once you get enough practice, you’d be able to identify
and draw forces in the main diagram itself instead of
making a separate one
Step 4 : Select axes and Write equations
When the body is in equillibrium then choose the axis
in such a manner that maximum number of force lie
along the axis.
If the body is moving with some acceleration then first
find out the direction of real acceleration and choose
the axis one is along the real acceleration direction
and other perpendicular to it.
Write the equations according to the newton’s second
law (Fnet = ma) in the corresponding axis.
EQUILIBRIUM OF CONCURRENT FORCE :
 0,
angle taken in order.
A
B
C
 0 or
can be represented completely by three sides of a tri-
B
Tb
net
Three concurrent forces will be in equilibrium, if they
A
(1)
F
0 ;
F
(6)
Lami's Theorem : For concurrent forces
F1
F
F
 2  3
sin  sin  sin 



SOLVED EXAMPLE
Example-1
Three forces starts acting simultaneously on a particle
moving with velocity

v. . These forces are represented
in magnitude and direction by the three sides of a triangle ABC (as shown). The particle will now move with
velocity
C
If all the forces working on a body are acting on the
same point, then they are said to be concurrent.
(2)
A body, under the action of concurrent forces, is said
to be in equilibrium, when there is no change in the
state of rest or of uniform motion along a straight
line.
6
A
B
Newton’s Laws of Motion and Friction
JEE - PHYSICS
(A)
v remaining unchanged
(B) Less than
v
(C) Greater than
(D)
Sol. :
tan  
 F  2F cos   0
v
 cos   1 / 2
v in the direction of the largest force BC
(A) Given three forces are in equilibrium i.e. net force
will be zero. It means the particle will move with same
velocity.
Example-2
Two forces are such that the sum of their magnitudes
is 18 N and their resultant is perpendicular to the smaller
force and magnitude of resultant is 12. Then the magnitudes of the forces are
(A) 12 N, 6 N
(B) 13 N, 5N
(C) 10 N, 8 N
(D) 16 N, 2 N
Sol.
2F sin 
 tan 90 = 
F  2 F cos 
(B) Let two forces are
F1
According to problem :
or
  120
Example-4
A mass M is suspended by a rope from a rigid support
at P as shown in the figure. Another rope is tied at the
end Q, and it is pulled horizontally with a force F. If the
rope PQ makes angle with the vertical then the tension in the string PQ is
P

F2 (F1  F2 ) .
and
F1  F2  18
...(i)
M
F1 and resultant (R) is 90°
F2 sin 

 tan 90 
F1  F2 cos 
R =12
(A)
F sin 
(B)
F / sin 
(C) F cos 
F2
(D) F / cos 
Sol. (B) From the figure
For horizontal equilibrium

F1
F1
F2
T sin   F
...(ii)
T
and R 2  F12  F22  2F1F2 cos 
144  F12  F22  2F1F2 cos 
F
Q
Angle between
 cos   
v
T cos

F
T sin
…..(iii)
mg
by solving (i), (ii) and (iii)
we get F1  5N and
F2  13N
 T 
Example-3
The resultant of two forces, one double the other in
magnitude, is perpendicular to the smaller of the two
forces. The angle between the two forces is
(A) 60 °
(B) 120°
(C) 150°
(D) 90°
Sol.
Let forces are F and 2F and angle between them is
and resultant makes an angle
 with the force F..
Newton’s Laws of Motion and Friction
F
sin 
Example-5
R=5m
A
B
3m
1m

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JEE - PHYSICS
Find out the normal reaction at point A and B if the
mass of sphere is 10 kg.
N1
N2
A
5m3m
53°
B
(c)
N1
53°
37°
O
x'
Now F.B.D.
x
10 g
y'
Fig-II
N = mg – F sin 
a=
mg
N = mg + F sin 
3N2
5
y
N1sin53° = 4N1/5
N1
N2
37°
4N2 O
N2cos37°=
5
53°
N1cos53° 
3N1
5
3N1
4N 2
=
...(1)
5
5
In y-direction
3N 2
4N1
+
= 100 ...(2)
5
5
after solving above equation
N1 = 80 N,
N2 = 60 N
MOTION OF A BLOCK ON A HORIZONTAL SMOOTH
SURFACE
(a) When subjected to a horizontal pull.
(b) When subjected to a pull acting at an angle  to the
horizontal.
8
A
a=
N
Fcos 
m
C
A
m

In x-direction
and
MOTION OF A BODY ON A SMOOTH INCLINED PLANE
(a)
When smooth inclined plane is fixed
(b)
When the smooth inclined plane is moving
horizontally with an acceleration b.
100
The body is in equilibrium so equate the force in x & y
direction
a
F
Now resolve the forces along x & y direction
N2sin37° 
N
F cos

F sin
N2
F cos
F
Fcos 
and a =
m
m
When subjected to a push acting at an angle  to the
horizontal.
and
y
F
mg
Fig-I
N = mg


m
F
mg
37°
1m 4m
F sin
a
m
O
Sol.
N
N


b
a
B

C
mg
Fig-I
N = mg cos 
and a = g sin 
N
m
a

B
mg
Fig-II
N = m (g cos  + b sin )
and a = (g sin  – b cos )
MOTION OF BODIES IN CONTACT
Force of contact :
(a)
Two bodies are kept in contact and force is applied
on the body of mass m1.
F
a= m m
1
2
and
f=
m2 F
m1  m 2
a
F
f
m1
f
m2
Newton’s Laws of Motion and Friction
JEE - PHYSICS
(b)
Two bodies are kept in contact and force is applied on
m1F
T1 = (m  m  m )
1
2
3
the body of mass m2
F
a= m m
1
2
and
f=
m1F
m1  m2
(m1  m2 )F
T2 = (m  m  m )
1
2
3
a
a
f
m1
(c)
f
m2
F
m1
Three bodies are kept in contact and force is applied
on the body of mass m1
a
T2
T1
T1
T2
a
m3F
T2 = (m  m  m )
1
2
3
Three bodies are kept in contact and force is applied
on the body of mass m3
a
m1
T1
m3
m2
T2
T1
F
T2
T
T
T
T

g

a
m2
m2g
m1g
When two bodies of masses m1 and m2 are attached at
the ends of a string passing over a pulley in such a way
that mass m1 rests on a smooth horizontal table and
mass m2 is hanging vertically
m2 g
a = (m  m )
1
2
F
m1F
a = m  m  m T1 = (m  m  m )
1
2
3
1
2
3
F
m3
 2m1m 2
T=  m m
 1
2
m1
(d)
T2
T2
When two bodies of mass m1 and m2 are attached at
the ends of a string passing over a pullley as shown in
the figure (neglecting the mass of pulley).
F
(m 2  m 3 )F
a = m  m  m T1 = (m  m  m )
1
2
3
1
2
3
(d)
m2
(m1 – m 2 )g
a = (m  m )
1
2
m3
m2
m1
F
(c)
T1
T1
m1m2g
T = (m  m )
1
2
motion
(m1  m2 )F
T2 = (m  m  m )
1
2
3
m1
a
T
MOTION OF CONNECTED BODIES
(a)
Two bodies are connected by a string and placed on a
smooth horizontal surface
F
a= m m
1
2
m1F
T= m m
1
2
m2
m2g
(e)
If in the above case, mass m1 is placed on a smooth
inclined plane making an angle  with horizontal as
shown in figure, then
a
m1
T
m2
F
When three bodies are connected through strings as
shown in fig. and placed on a smooth horizontal surface.
a=
(m 2 – m1 sin )g
m1  m 2
n
io
ot
m
m1
F
a = (m  m  m )
1
2
3

Newton’s Laws of Motion and Friction
T=
m1m 2 g(1  sin )
m1  m 2
motion
(b)
T
T
T
m2
m2g
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JEE - PHYSICS
If the system remains in equilibrium,
then m1g sin = m2g
(f)
If case (d), masses m1 and m2 are placed on inclined
planes making angles  and  with the horizontal respectively, then.
 If m1sin > m2sin
g(m1 sin  – m 2 sin )
then a =
(m1  m 2 )
on
m1m 2
T = m  m (sin + sin) g
1
2
mo
ti
T
T
APPARENT WEIGHT OF A BODY IN A LIFT
The lift is at rest or moving with uniform velocity
a=0:
a=0
v=constant
mg – N = 0 or
N = mg
or
Wapp.= W0
Wapp. = N = reaction of supporting surface
and
W0 = mg = true weight.
When the lift moves upwards with an acceleration a :
m
ot
io
n
a
m2
m1


SOLVED EXAMPLE
Example-6
A 5 kg block has a rope of mass 2 kg attached to its
underside and a 3 kg block
F0
is suspended from the other end of the
5 kg
rope. The whole system is accelerated
upward is 2 m/s2 by an external force F0.
2 kg
(a) What is F0?
3 kg
(b) What is the force on rope?
(c) What is the tension at middle point of the rope?
(g = 10 m/s2)
Sol.
For calculating the value of F0, consider two blocks
with the rope as a system.
F.B.D. of whole system
F0
(a)
10kg
2m/s
2
10 g = 100N
F0 – 100 = 10 × 2
F = 120 N
...(1)
(b) According to Newton’s second law, net force on
rope.
F = ma = (2) (2) = 4 N ...(2)
(c) For calculating tension at the middle point we
draw F.B.D. of 3 kg block with half of the rope
(mass 1 kg) as shown.
T – 4g = 4 × (2) = 48 N
T
or N = m (g + a) = mg  1  a 
 g


N – mg = ma
 a
Wapp. = W0  1  g 



When the lift moves downwards with an acceleration a :
a
mg – N = ma
or
a
N = m(g – a) = mg  1 – 
 g
 a
Wapp. = W0  1 – 
 g
Here, if a> g, Wapp. will be negative. Negative apparent
weight will mean that the body is pressed against the
roof of the lift instead of floor.
When the lift falls freely, i.e., a = g :

a=g
N = m(g – g) = 0 or
Wapp. = 0
MONKEY CLIMBING A ROPE
The tension in the rope = T
a=0
a
a
a
v=const.
4g
10
T = mg
T = m(g + a)
T = m(g – a)
Newton’s Laws of Motion and Friction
JEE - PHYSICS
SYSTEMS OF VARIABLE MASS


According to Newton’s 2nd law : F  m
d v  dm
v
dt
dt
If the mass of the system is variable and the velocity

of escaping mass is constant


Fv
dv
= 0
dt
and
dm
dt
Bullet fired from a machine gun :
CONSTRAINED MOTION:
String Constraint :
When two objects are connected through a string and
if the string have the following properties :
(a) The length of the string remains constant i.e.
inextensible string.
(b) Always remains tight, does not slacks.
Then the parameters of the motion of the objects along
the length of the string and in the direction of extension
have a definite relation between them.
(i)
1st format : - When string is fixed
s
A
F
If 'n’ bullets each of mass m are fired per unit time
from a machine gun.The force required to hold the gun
B
v
The block B moves with velocity v. i.e. each particle of
block B moves with velocity v.
If string remain attached to block B it is necessary that
velocity of each particle of string is same = v (vs = v)
Now we can say that Block A also moves with velocity v.
dm
= v(mn) = mnv..
dt
Sand is dropped on to a conveyor belt :
If sand is dropped on a conveyor belt at a constant
rate,
= v×
v
v
A
B
vA = vB = v
If pulley is fixed then the velocity of all the particles of
string is same along the string.
extra force required to maintain the constant speed
of the belt = v
dm
dt
SOLVED EXAMPLE
Jet of liquid coming out from a pipe :
If a jet of liquid coming out from a pipe of uniform
area of cross-section A hits a wall normally with a
velocity v.
The force applied by the jet on the wall = Av2
( = density of liquid)
Rocket :
Let the rocket ejecting gases, with a constant velocity
= vrel. (relative to rocket)
By Newton’s 2nd law : Fext. + vrel.  dm  = ma
 dt 
dm
(the reaction force
thrust on the rocket = vrel
dt
exerted by the leaving gases on the rocket)
In the absence of air resistance, if the rocket moves up
with an acceleration = a
thrust = vrel.
dm
= ma + mg
dt
If the rocket just moves up with constant velocity then
thrust = vrel
Example-7
v
B
A vA =?
Sol.
In the above situation block B is moving with velocity
v. Then speed of each point of the string is v along the
string.
 speed of the block A is also v
v
B
v
A vA=v
dm
= mg
dt
Newton’s Laws of Motion and Friction
11
JEE - PHYSICS
Example-8
v
A
VA = 8 m/s
37°
vB
37°
B
vB=?
Sol.
along string is v then
vBcos37° = v
 Block A is moving with velocity 8 ms–1.
along the string.
break only when the compoent of vB along string is
Example-10
8 m/s.
8 m/s
..(2)
from (1) & (2) vB cos37° = 10 cos53°
10  3 / 5
30 15
 m / sec
 vB =
=
4/5
4
2
 velocity of every point on the string must be 8m/s
The real velocity of B is vB. Then the string will not
B
50/3 m/s
53°
A
8 m/s
8 m/s
37°
A
B
vB

8
vB cos 37° = 8  vB =
= 10 m/sec
cos 37
Sol.
Example-9
Find out the velocity of block B in a pulley block
system as shown in figure.
(ii)
10 m/s
Sol.
37°
53°
B
A
Let us assume the velocity of all the particle of string is
v.
What is the velocity of block A in the figure as shown
above.
The component of velocity of ring along string
= velocity of A
50
cos 53 = v  v = 10 m/s
=
A
A
3
In the first format only two points of string are attached
or touched to moving bodies.
2nd format When pulley is also moving
To understand this format we consider the following
example in which pulley is moving with velocity vp
and both block have velocity vA & vB respectively as
shwon in figure.
If we observe the motion of A and B with respect to
pulley. Then the pulley is at rest. Then from first
format.
vAP = – vBP
vP
v
v
53°
10m/s
37°
A
53°
B
10cos53°
10 m/s is the real velocity of block A then its component
along string is v.

10 cos 53° = v
...(1)
If vB is the real velocity of
block B then it component
12
vA
vB
A
B
(–ve sign indicate the direction of each block is
opposite with respect to Pulley)
vA  vB
vA – vp = – vB + vP  vP =
2
To solve the problem put the values of vA, vB, & vP
with sign.
Newton’s Laws of Motion and Friction
JEE - PHYSICS
Now
10 m/s
 7=
Example-11
vP
Now
A v=?
A
vP =
–3 / 2  v F
31
 14 + 3/2 = vF  vF =
2
2
vC  vD
8  vD
31
=
= vF 
2
2
2
 vD = 31 – 8
vD = 23 m/s (upward direction)
Example-14
B
Sol.
v E  vF
= 7 m/s
2
SOLVED EXAMPLE
vA  vB
2
C
B
Putting vp = 10 ms–1, vB = 0,
D
we get
E
vA = 20 ms–1 (upward direction)
A H
Example-12
10 m/sec
vP= 10m/s
Sol.
5m/s
Sol.
B v=?
B
A
+ 10 =
Example-13
Now
7 m/s
8m/s
F
Sol.
(iii)
D
5m/s
A
B C
Find out the velocity of Block D
From 2nd format of constrained motion
vA  vB
vE =
2
2–5
= –3/2
vE =
2
(If upward direction is taken to be +ve)
vE = –3/2 m/s
Newton’s Laws of Motion and Friction
vH  vE
2
vH = 10 m/s  if upward direction is taken to be positive
–5  v B
2
vB = 25 m/sec (in upward direction)
10 =
2m/s
Find the velocity of point G.
In string ABCD from first format of constrain
VD = 10 m/s
Now vD =
If we take upward direction as +ve then
E
F
G
m
10  v E
 vE = 30 m/s 
2
v F  vG
–10  vG
= vE  30 =
2
2
60 + 10 = vG
vG = 70 m/s
In IInd format three or four Points of the string is
attached to the moving bodies.
3rd format of string constraint :
SOLVING STRATEGY :
1.
First choose the longest string in the given problem
which contains the point of which velocity/
acceleration to be find out.
2.
Now mark a point on the string wherever it comes in
contact or leaves the contact of real bodies.
3.
If due to motion of a point, length of the part of a string
with point is related, increases then its speed will be
taken +ve otherwise –ve.
13
JEE - PHYSICS
SOLVED EXAMPLE
4m/s
c
b
Example-15
e
x
a
A
I
E H
J
C
B
C F
5m/s
A
8m / s  A
2m/s
Step 1. We choose a longest string ABCDEFGHIJ in which we
=
have to find out velocity of point J (vc)
Step 2. Mark all the point A, B ................
Step 3. Write equation
vA + vB + vC + vD + vE + vF + vG + vH + vI + vJ = 0
objects)
vB = vC = 5 m/s
(increases the length)
vF = vG = 2m/s
(It also increases the length)
Let us assume C is moving upward with velocity vc so
vc negative because it decreasing the length
vC = 14 m/sec (upward)
Example-16
4m/s
F
c  2m / sec
E
D
Find out the velocity of block E as shown in figure.
14
vA  vB
(from second format)
2
vK  vC
(from 2nd format)
2
from 2nd format of constrain vx =
vB + vc + vF + vG + vJ = 0 ...(1)
1 m / s
z
=
(No movement of that point because attached to fixed
B  2m / s
D
3 2
= 5/2 m/sec. (upward)
2
vx = 4m/s (from first format of constrain)
vA = vD = vE = vH = vI = 0
8m / s  A
j
E
8–2
= 3 m/sec. (upward)
2
va =
5 + 5 + 2 + 2 – vc = 0
i
Step 2 : Now write equation according to the velocity of each
point (either increase or decrease the length)
va + vb + vc + vd + ve + vf + vg + vh + vi + vj= 0 ...(1)
Now find value of va, vb ..... in a following way
vk =

g
f
1m / s 
Sol.
Sol.
Step-1
B  2m / s
vC=?
G
B
h y
c  2m / s
k
D

F
d
We first choose the longest string in which point j (block
E) lie. (abcdefghij)
(iv)
v y  vz
2
 vz = 0 (fixed)
 vy = 2 vx = 8 m/s (upward)
 Now va = – 5/2 m/s (decreases the length)
vb = vc = vd = ve = 0 (attached to fixed object)
vf = vg = 1m/s (increases the length)
vh = vi = vy = 8 m/s (increase the length)
Let us assume block E move upward then vj = – vE
(decrease the length)
Puting the above values in eq. (1)
 –5/2 + 1 + 1 + 8 + 8 – vE = 0
vE = 31/2 m/s (upward)
Steps to solve more complex problem of
String Constraint :
Step 1. Identify all the objects and number of strings
in the problem.
Step 2. Assume variable to represent the parameters
of motion such as displacement, velocity acceleration etc.
(i) Object which moves along a line can be specified by one variable.
(ii) Object moving in a plane are specified by two
variables.
(iii) Objects moving in 3-D requires three variables to
Newton’s Laws of Motion and Friction
JEE - PHYSICS
represent the motion.
Step 3. Identify a single string and divide it into different linear sections and write in the equation format.
 1+  2+  3+  4+  5+  6 = 
Step 4. Differentiate with respect to time
d1 d 2 d 3


 ....  0
dt
dt
dt
direction. If the values of a1 and a2 come out to be
positive then only the assumed directions are correct;
otherwise the body moves in the opposite direction.
Since the pulley is smooth and massless, therefore, the
tension on each side of the pulley is same.
The free body diagram of each block is shown in the
figure.
d1
= represents the rate of increment of the portion 1,
dt
F.B.D. of m2
F.B.D. of m1
end points are always in contact with some object so
take the velocity of the object along the length of the
string
d1
 V1  V2
dt
Take positive sign if it tends to increase the length and
negative sign if it tends to decrease the length. Here
+V1 represents that upper end is tending to increase
the length at rate V1 and lower end is tending to increase the length at rate V2.
Step 5. Repeat all above steps for different-different
strings.Let us consider a problem given below
Here 1 + 2 = constant
Applying Newton’s second Law on blocks m1 and m2
Block m1
m1g – T = m1a1
.............(1)
–m2g + T= m2a2
.............(2)
Block m2
Number of unknowns : T, a1 and a2 (three)
d 2
d1
0
+
dt
dt
Number of equations: only two
(V1 – VP) + (VP – V2) = 0
Obviously, we require one more equation to solve the
V1  V2
2
Similarly,
problem. Note that whenever one finds the number of
Vp =
aP =
a1  a 2
2
equations less than the number of unknowns, one must
think about the constraint relation. Now we are going
Remember this result
SOLVED EXAMPLE
Example-17
Two blocks of masses m1 and m2 are attached at the
ends of an inextensible string which passes over a
smooth massless pulley. If m1 > m2, find :
(i) the acceleration of each block
(ii) the tension in the string.
to explain the mathematical procedure for this.
How to determine Constraint Relation ?
(1) Assume the direction of acceleration of each
block, e.g. a1 (downward) and a2 (upward) in this
case.
(2) Locate the position of each block from a fixed
point (depending on convenience), e.g. centre of
the pulley in this case.
(3) Identify the constraint and write down the equation of constraint in terms of the distance assumed. For example, in the chosen problem, the
length of string remains constant is the constraint
or restriction.
Thus, x1 + x2 = constant
Sol.
The block m1 is assumed to be moving downward and
the block m2 is assumed to be moving upward. It is
merely an assumption and it does not imply the real
Newton’s Laws of Motion and Friction
Differentiating both the sides w.r.t. time we get
dx1
dx 2
+
=0
dt
dt
Each term on the left side represents the velocity of the
blocks.
15
JEE - PHYSICS
Since we have to find a relation between accelerations,
therefore we differentiate it once again w.r.t. time.
Thus
2
d x1
2
+
d x2
2
=0
d 2 x1
= + a1
dt 2
and block m2 is assumed to be moving upward (x2 is
decreasing with time)

d2 x 2
2
F.B.D. m1
F.B.D.
of pulley
 m1  m 2 
a = m  m  g
 1
2
Block m1 : m1g – T1 = m1a1
....(1)
Block m2 : m2g – T1 = m2a2 ....(2)
Block m3 : m3g – T2 = m3a3 ....(3)
Pulley : T2 = 2T1
....(4)
Number of unknowns a1, a2, a3, T1 and T2 (Five)
Number of equations : Four
The constraint relation among accelerations can be
= – a2
dt
Thus a1 – a2 = 0
or a1 = a2 = a (say) is the required constraint relation.
Substituting a1 = a2 = a in equations (1) and (2) and
solving them, we get
(i)
F.B.D. m2
Applying Newton’s Second Law to
2
dt
dt
Since, the block m1 is assumed to be moving downward (x1 is increasing with time)

F.B.D. m3
 2m1m 2 
(ii) T =  m  m  g
 1
2
Example-18
A system of three masses m1, m2 and m3 are shown in
the figure. The pulleys are smooth and massless; the
strings are massless and inextensible.
(i) Find the tensions in the strings.
(ii) Find the acceleration of each mass.
obtained as follows
For upper string
x3 + x0 = c1
For lower string
x2 – x0) + (x1 – x0) = c2
x2 + x1 – 2x0 = c2
Eliminating x0 from
the above two relations,
we get x1 + x1 + 2x3 = 2c1 + c2 = constant.
Differentiating twice with respect to time,
we get
d 2 x1
dt 2
+
d2 x 2
dt 2
or a1 + a2 + 2a3 = 0
+2
d2 x3
dt 2
=0
......(5)
Solving equations (1) to (5), we get
(i)


4m1m 2 m3
T1 =  4m m  m (m  m )  g ; T2 = 2T1
3
1
2 
 1 2
 4m1m 2  m1m3  3m 2 m3 
(ii) a1 =  4m m  m (m  m )  g ;

1 2
3
1
2 
 3m1m 3  m 2 m 3  4m1m 2 
a2 =  4m m  m (m  m )  g
1 2
3
1
2


Sol.
All the blocks are assumed to be moving downward
and the free body diagram of each block is shown in
figure.
16
 4m1m 2  m 3 (m1  m 2 ) 
a3 =  4m m  m (m  m )  g
3
1
2 
 1 2
Newton’s Laws of Motion and Friction
JEE - PHYSICS
WEDGE CONSTRAINT :
Conditions :
(i) There is a regular contact between two objects.
(ii) Objects are rigid.
The relative velocity perpendicular to the contact plane
of the two rigid objects is always zero if there is a
regular contact between the objects. Wedge constraint
is applied for each contact.
In other words,
Components of velocity along perpendicular direction
to the contact plane of the two objects is always equal
if there is no deformations and they remain in contact.
SOLVED EXAMPLE
SPRING FORCE
Many spring follow Hooke's law for small extension
and compression. That is, the extension or compression – the increase or decrease in length from the
relaxed length – is proportional to the force applied to
the ends of the spring.
Hook's Law for an ideal spring F = kL
In equation F is the magnitude of the force exerted on
each end of the spring and L is the modulus of change
in length of the spring from its relaxed length. The
constant k is called the spring constant for a particular spring. The SI units of a spring constant are N/m.
When we say an ideal spring, we mean a spring that
obeys Hooke's law and is also massless. Since we have
assumed spring to be massless we know forces acting
on both ends have to be equal and opposite, to have
net force on spring to be zero.
NOTE :
If we look at FBD of the spring we will not that force
on spring must act from both ends.
Example-19
A rod of mass 2m moves vertically downward on the
surface of wedge of mass as shown in figure. Find
the relation between velocity of rod and that of the
wedge at any instant.
F
F
l0 + x
Lets say Rahul and Sachin are pulling a spring from
two ends as shown. Rahul moves x2 and Sachin moves
x1.
Rahul
l0
x2
x1
Sol.
Using wedge constraint.
Component of velocity of rod along perpendicular to
inclined surface is equal to velocity of wedge along
that direction.
Sachin
The force acting on Rahul and Sachin is k(x1 + x2),
Not kx2 on Rahul and kx1 on Sachin. Force due to spring
is kx where x is defined as |l – l0| , where l is present
length and l0 is natural length.
EQUIVALENT SPRING CONSTANT.
(a)
When springs are connected in parallel then we can
replace them by single spring of spring constant ke
where ke = k1 + k2.
k1
k2
u cos  = v sin 
u
= tan u = v tan 
v
Newton’s Laws of Motion and Friction
k3
k4
17
JEE - PHYSICS
(b)
For more than two spring k = k1 + k2 + k3 + .............
When springs are connected in series then we can
replaced them by single spring of spring constant
ke where 1/ke = 1/k1 + 1/k2 . As spring constants are not
equal so extensions will not be equal , but total extension
y can be written as sum of two extensions y = y1 + y2
for more than two springs.
k1
k2
k3
kx1
m
kx1– mg = ma3  a3 = 5g 
mg
k4
2m
1 1 1
 
 ............
k k1 k 2
kx3
kx3 + 2mg = 2ma2  a2 =
2mg
SOLVED EXAMPLE
Example-20
The system shown in the figure is in equilibrium. Find
the initial acceleration of A,B, and C just after the
spring-2 is cut.
5g
2 
kx3
3m
spring 1
acceleration of 3m will be zero.
k
3mg
A
spring 2
k
g
IMPORTANT POINT
B
spring 3
It is important to remember that ropes can change
k
tension instantaneously while spring need to move to
change tension, so in this example tension in spring
C
x1 k
x2
k
x3
k
kx3
is not changing instantaneously
kx2
kx1
3m
2m
m
3mg
kx3 + 2mg
kx2 + mg
SOLVED EXAMPLE
Example-21
In following setup pully strings and spring are light.
initially all masses are in equilibrium and at rest.
Sol.
C
S1
3m
S2
3mg = kx3 .....(1)
2mg + kx3 = kx2
2mg + 3mg = kx2  5mg = kx2 ....(2)
kx1 = 6m
.....(3)
when spring 2 is cut spring force in other two strings
remain unchaged.
18
B m
S1
A 4m
(a) find tension in spring and tension in ropes
Newton’s Laws of Motion and Friction
JEE - PHYSICS
(b) find accleration of masses immediately after the
string S3 is cut.
C
S1
3m
S2
Sol.
B m
(a)
S1
A 4m
T1
T2
m
4m
4mg
mg
3m
T2
T1
T1
applying Newton's 2nd law to block A 4mg – T1 = 0
applying Newton's 2nd law to block B mg – T1 – T2
=0
applying Newton's 2nd law to block C T3 – T2 = 0
Solving T1 = 4 mg
T2 = 5 mg
T3 = 5mg
Here spring is behaving same as string except that it
is streched while string can not strech.
[b] The most important point in this problem is that
any object of finite mass can not charge its position
instantaneously as this require infinite velocity.
Thus immediately after cutting the string S3 all masses
will reamin at same position and force due to spring
will not change. As force of srping is kx. At the same
time we would like to emphasize that tension in string
S 2 will change instataneously (Tension is a self
adjsuting force). To maintain constraint relation
blocks B and C have same since they magnitude of
acceleration. We can identify all the forces acting on
all objects. only tension T in string S2 is unknown force
all other forces are known.
considering FBD
T'2
T1
We know from part [a] that tension in sprng is T1 and
T1 = 4 mg
Writing Newton's Second law for A
4mg – T1 = 4maa'
Writing Newton's Second law for B
T1 + mg – T2 = maa'
Writing Newton's Second law for C
T2' = 3maC'
Qunatities which may have different value from part
[a] are represented using symbol, for eg a tensen in
string s2 is T2', others which have same value as part [a]
have been retained with same symbol.
as
aB' = aC'
5
solving we get
aA' = 0
aB' = aC' = g
4
PSEUDO FORCE
Motion in Accelerated Frames :
Till now we have restricted ourselves to apply
Netwon's laws of motion, only to describe observation that are made in an inertial frame of reference. In
this part, we learn how Netwon's laws can be applied
by an observer in a noninertial reference frame. For
example consider a block kept on smooth surface of
a compartment of train.
If the train accelerates, the block accelerates toward
the back of the train. We may conclude based on
Newton's second law F = ma that a force is acting the
block to cause it to accelerate, but the Newton's second law is not applicable from this non - inertial frame.
So we can not relate observed acceleration with the
force acting on the block.
If we still want to use Newton's second law we need
to apply a pseudo force, acting in backward direction,
i.e. opposite to the acceleration of noninertial reference frame. This force explains the motion of block
towards the back of car. The frictions force is equal
to – ma1 where a is the acceleration of the noninertial
reference frame. Fictitious force appears to act on an
object in the same way as a real force, but real forces
are always interactions between two objects. On the
other hand there is no second object for a fictitious
force.
a'C
4m
4mg
a'A a'
B
m
3m
mg
T1
Newton’s Laws of Motion and Friction
SOLVED EXAMPLE
Example-22
A small ball of mass m hangs by a cord from the ceiling
of a compartment of a train that is accelerating to the
right as shown in figure (a). Analyze the situation for
two observes A and B.
19
JEE - PHYSICS
Sol.
The observer A on the ground, in inertial frame. He
sees the compartment is accelerating and knows that
the deviation of the cord provides the ball, required
horizontal force. The noninertial observer on the compartment, can not see that car's motion so that he is not
aware of its acceleration. Because he does not know of
this acceleration, he will say that Newtons' second law
is not valid as the object has net horizontal force [the
horizontal component of tension] but no horizontal
acceleration.
a0

T
Inertial
observer (A)


mg
Figure (a)
Non Inertial observer (B)


Ffictitious
T


mg
Figure (b)
for the inertial observer , ball has a net force in the
horizontal direction and is in equilibrium in the vertical
direction. For the non-inertial observer, we apply
fictitious force towards left and consider it to be in
equilibrium.
According to the inertial observer A, the ball
experience two forces, T exerted by the cord and the
weight.
Apply Newton's second law in horizontals and vertical
direction we get.
Inertial observer Tsin – ma = 0
Tcos – mg = 0
According the noninertial observer B riding in the car
[fig. b], the ball is always at rest and so its acceleration
is zero. The noninertial observer applies a fictitious
force in the horizontal direction of magnitude ma
toward left. This fictitious force balances the
horizontal component of T and thus the net force on
the ball is zero.
Apply Newton's second law in horizontal and vertical
direction we get.
Noninertial observer
Tsin – F = 0
Tcos – mg = 0
20
The noninertial observer B obtaines the same equations
as the intertial observer. The physical explanation of
the cord's deflection, however, differs in the two frames
of reference.
SOLVED EXAMPLE
Example-23
Apparent weight in an Elevator
A passenger weighing 600 N rides in an elevetor. What
is the apparent weight of the passenger in each of the
following situations? In each case, the magnitude of
the elevator's acceleration is 0.500 m/s2. [a] The
passenger is on the first floor and has pushed the
button for the 15th floor; the elevator is beginning to
move upward . [b] the elevator is slowing down as it
nears the 15th floor.
Sol.
Lets understand this situation by making FBD of the
person inside the elevator.
Please note that ma is not being shown in FBD. as ma
is not the force, it is effect of the net force acting on
the person. Only two actual forces act on the person
in elevator normal contact force by the scale's surface
and weight. By writing Newton's 2nd law we can find
the normal force from the known weight and the
acceleration. W = 600 N ; magnitude of the acceleration
is a = 0.500 m/s2 . To find : W'
We expect the apparent weight W' = N to be greater
then the true weight - the floor must push up with a
force greater then W to cause an upward acceleration.

y

a
N
Vector sum of forces

N
mg
m

F
mg
Free body
diagram
[a]
[b]
   

F = N + mg = ma
F is upward so
N > mg
[c]
N – W = may
Since W = mg, we can substitute m = W/g.
 ay 
W
N = W + ma = W + g a y  W 1  g 


2
 0.500m / s 
= 600N  1  10m / s 2   630 N


Newton’s Laws of Motion and Friction
JEE - PHYSICS
(b)
By Newton's third law same force will be acting on the
scale. Hence scale will measure your weight larger than
the actual weight.
When the elevator approaches the 15th floor, it is
sloving down while still moving upward; its
acceleration is downward (ay < 0) as in figure.
y

N
Vector sum of forces

N

a
mg
F
Free body
diagram
[a]
mg

[b]
Kinetic Friction (fk)
It acts on the two contact surfaces only when there
is relative sliping or relative motion between two
contact surfaces.
fk = kN
The relative motion of a contact surface with respect
to each other is opposed by a force given by
fk = k N, where N is the normal force between the
contact surfaces and  k is a constant called
‘coefficient of kinetic friction’, which depends,
largely, on the nature of the contact surfaces.
N
   

[c]
 a
N  W 1  
 g
 0.500m / s 2 
 600N  1 
  570 N
10m / s2 

FRICTION
When an object is in motion either on a surface or
in a viscous medium such as air or water, there is
resistance to the motion because the object interacts
with its surroundings. We call such resistance a force
of friction. Forces of friction are very important in our
everday lives. They allow us to walk or run and are
necesary for the motion of wheeled vehicles.
Friction force is of two types :
(1) Static frictions ‘fs’
(2) Kinetic friction ‘fk’
(1)
Static Friction :
The static friction between two contact surfaces is
given by fs < s N, where N is the normal force
between the contact surfaces and s is a constant
which depends on the nature of the surfaces and is
called ‘coefficient of limiting friction’.
Static friction acts on stationary objects. Its values
satisfy the condition
fs < s N
s = Co-efficient of limiting friction
Static friction takes its peak value (fs(max) = sN)
when one surface is ‘about to slide’ on the other.
Static friction in this case is called limiting friction.
v
 kN
F = N + mg = ma
F is upward so
N < mg
The normal force must be less than the weight to have
an downward net force.
Newton’s Laws of Motion and Friction
(2)
mg
NOTE :
(1)
Value of k is always less than s (k < s) from
experimental observation.
(2)
If only coefficient of friction () is given by a problem
then use s = k = 
(3)
Value of s and k is independent of surface area it
depends only on surface properties of contact surface.
(4)
k is independent of relative speed.
(5)
s & k are properties of a given pair of surfaces i.e. for
wood to wood combination 1 , then for wood to iron
2 and so on.
Static
friction
Friction
force
Kinetic
friction
45º
fs (max)
Applied
force
Graphical representation of variation of friction force
with the force applied on a body
SOLVED EXAMPLE
Example-24
Find frictional force acting on the block in following
cases given that  = 0.1
100N
F1
10 kg
F2
21
JEE - PHYSICS
Sol.
Case - 1
F2 = 5 N
F2 = 15 N
F2 = 25 N
Case - 2
F2 = 10 N
F2 = 20 N
F2 = 42 N
F2 = 50 N
ANGLE OF FRICTION
The angle of friction is the angle which the resultant
of limiting friction FL and
F1 = 0
fr = 5 N (towards left)
fr = 15 N (towards left)
fr = 20 N (towards left)
F1 = 20 N
fr = 10 N (towards right)
fr = 0
fr = 20 N (towards left)
fr = 20 N (towards left)
R
R = FL+ N

F
FL
Example-25
Find friction force acting on block in the following
cases.
Case - I
N
F1 = 0 , F2 =
5 2
Case - II F1 = 0 , F2 = 50 2
Case - III F1 = 5N , F2 = 5 2
W = mg
normal reaction N makes with the normal reaction. It is
represented by ,
Thus from the figure. tan  
FL
N
or
tan  = µ
( FL = µ N)
For smooth surfaces,  = 0 (zero)
F2

F1
10 kg
SOLVED EXAMPLE
µ=0.1
F2

10 kg
F1
Sol.
10 kg
F1
µ=0.1
µ=0.1
Both of the above are same cases.
Case I
45°
F2
Example-26
A body of mass 400 g slides on a rough horizontal
surface. If the frictional force is 3.0 N, find (a) the
angle made by the contact force on the body with the
vertical and (b) the magnitude of the contact force.
Take g = 10 m/s2.
Sol.
Let the contact force on the block by the surface be Fc
which makes an angle  with the vertical (shown figure)
Fc
N
f
5 N = 105 N
fr
100
The component of Fc perpendicular to the contact
surface is the normal force N and the component of F
parallel to the surface is the firction f. As the surface is
horizontal, N is vertically upward. For vertical
equilirbrium,
N = Mg = (0.400 kg) (10 m/s2) = 4.0 N
The frictional force is f = 3.0 N
f
3
(a) tan    or,  = tan–1 (3/4) = 37º
N 4
(b) The magnituded of the contact force is
5
frmax = 0.1 × 105 = 10.5 N
fr = 5N
Case II
N
50 N = 150 N
fr
50 100
frmax = 0.1 × 150 = 15 N
so block will skid and fr = 15 N
N
F  N2  f 2
=
case - III
fr = 0
22
5N
5N
105
N
(4.0 N) 2  (3.0N) 2  5.0 N
MOTION ON A ROUGH INCLINED PLANE
Suppose a motion up the plane takes place under the
action of pull P acting parallel to the plane
Newton’s Laws of Motion and Friction
JEE - PHYSICS
N = mg cos 
Frictional force acting down the plane
F = N =  mg cos 
Appling Newton’s second law for motion up the plane
N
f
P
N
mg sin 
f 
mg
mg cos 
Where f is force of static friction on the block. For
normal direction to the plane, we have N=mg cos
As  increases, the force of gravity down the plane,
mg sin , increases. Friction force resists the slide
till it attains its maximum value.
fmax = N =  mg cos 
Which decreases with  (because cos  decreases as 
increases)
Hence, beyond a critical value  = c, the blocks starts
to slide down the plane. The critical angle is the one
when mg sin  is just equal of fmax, i.e., when
mg sin c =  mg cosc
or tan c = 
where C is called angle of repose
If  > c, block will slide down.For  < c the block
stays at rest on the incline.
P – (mg sin  + f) = ma
P – mg sin  –  mg cos  = ma
If P = 0 the block may slide downwards with an
acceleration a. The frictional force would then act up
the plane
mg sin  – F = ma
or,
mg sin  –  mg cos  = ma
SOLVED EXAMPLE
Example-27
A 20 kg box is gently placed on a rough inclined plane
of inclination 30° with horizontal. The coefficient of
sliding friction between the box and the plane is 0.4.
Find the acceleration of the box down the incline.
Y
F  N
mg
O
mgcos
X'
Sol.
X
Example-28
If µ = 0.9 and  = 45°, then find tension in string
Y'
In solving inclined plane problems, the X and Y
directions along which the forces are to be considered,
may be taken as shown. The components of weight of
the box are
(i) mg sin  acting down the plane and
(ii) mg cos  acting perpendicular to the plane.
N = mg cos 
mg sin  –  N = ma  mg sin  –  mg cos  = ma
a = g sin  – g cos 
= g (sin  –  cos )
1
3
= 9.8  2 – 0.4  2  = 4.9 × 0.3072 = 1.505 m/s2


The box accelerates down the plane at 1.505 m/s2.
ANGLE OF REPOSE :
Consider a rough inclined plane whose angle of
inclination  with ground can be changed. A block of
mass m is resting on the plane. Coefficient of (static)
friction between the block and plane is .
For a given angle , the FBD (Free body diagram) of
the block is
Newton’s Laws of Motion and Friction
SOLVED EXAMPLE
m
N
mgsin
mg cos
mg sin

Sol.
As < c block has no tendency to move hence tension
in the string will be zero.
TWO BLOCK PROBLEMS
Example-29
Find the acceleration of the two blocks. The system is
initially at rest and the friction coefficient are as shown
in the figure?
µ=0.5
Smooth
10 A
F = 50 N
10 B
//////////////////////////////////////////////////
Sol.
Method of solving
Step 1 : Make force diagram.
Step 2 : Show static friction force by f because value of
friction is not known.
23
JEE - PHYSICS
Step 3 : Calculate separately for two cases.
Case 1 : Move together
Step 4 : Calculate acceleration.
Step 5 : Check value of friction for above case.
Step 6 : If required friction is less than available it means
they will move together else move separately.
Step 7: (a) above acceleration will be common
acceleration for both
Case 2 : Move separately
Step 7(b) If they move separately then kinetic friction
is involved. whose value is µN.
Step 8 : Calculate acceleration for above case.
f
A
50
B
f
fmax = µN
 f  50 N (available friction)
50
= 2.5 m/s2
10  10
(i)
a=
(ii)
Check friction for B : f = 10 × 2.5 = 25
frmax = µN = 0.5 × 150 = 50 N
25 N is required which is less than available friction
hence they will move together.
and aA = aB = 2.5 m/s2
24
Example-30
Find the acceleration of the two blocks. The system is
initially at rest and the friction coefficient are as shown
in the figure?
µ=0.5
smooth
101 N
10 A
10 B
//////////////////////////////////////////////////
Sol.
fmax = 50 N
f
 f  50 N
101
A
f
B
(i) If they move together a =
(ii) Check friction on B
101
= 5.05 m/s2
20
f
10
f = 10 × 5.05 = 50.5 (required)
50.5 > 50 (therefore required > available)
Hence they will not move together.
(iii) Hence they move separately so kinetic friction is
involved.
fk=50
A
101
B
fk = µN=50
101  50
50
= 5.1 m/s2  aB =
= 5 m/s2
10
10
Also aA > aB as force is applied on A.
 for aA =
Newton’s Laws of Motion and Friction
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MODULE - 02
Newton’s Laws of Motion and Friction
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JEE - PHYSICS
EXERCISE-I
Scan for Video Solution
NEWTON’S LAWS OF MOTION
Laws of motion : Theoritical discription
Q.1
A rider on horse back falls when horse starts running
all of a sudden because
(1) Rider is taken back
(2) Rider is suddenly afraid of falling
(3) Inertia of rest keeps the upper part of body at rest
whereas lower part of the body moves forward with
the horse
(4) None of the above
Q.2
Q.3
Q.4
When a train stops suddenly, passengers in the
running train feel an instant jerk in the forward direction
because
(1) The back of seat suddenly pushes the passengers
forward
(2) Inertia of rest stops the train and takes the body
forward
(3) Upper part of the body continues to be in the state
of motion whereas the lower part of the body in
contact with seat remains at rest
(4) Nothing can be said due to insufficient data
A man getting down a running bus falls forward because
(1) Due to inertia of rest, road is left behind and man
reaches forward
(2) Due to inertia of motion upper part of body
continues to be in motion in forward direction while
feet come to rest as soon as they touch the road
(3) He leans forward as a matter of habit
(4) Of the combined effect of all the three factors stated
in (1), (2) and (3)
A boy sitting on the topmost berth in the compartment
of a train which is just going to stop on a railway station,
drops an apple aiming at the open hand of his brother
sitting vertically below his hands at a distance of about
2 meter. The apple will fall
(1) Precisely on the hand of his brother
(2) Slightly away from the hand of his brother in the
direction of motion of the train
(3) Slightly away from the hand of his brother in the
direction opposite to the direction of motion of the
train
(4) None of the above
Newton’s Laws of Motion and Friction
Q.5
A force of 100 dynes acts on a mass of 5 gm for 10 sec.
The velocity produced is
(1) 2 cm/sec
(2) 20 cm/sec
(3) 200 cm/sec
(4) 2000 cm/sec
Q.6
A machine gun is mounted on a 2000 kg car on a
horizontal frictionless surface. At some instant the gun
fires bullets of mass 10 gm with a velocity of 500 m/sec
with respect to the car. The number of bullets fired per
second is ten. The average thrust on the system is
(1) 550 N
(2) 50 N
(3) 250 N
(4) 250 dyne
Q.7
A particle of mass 0.3 kg is subjected to a force F = –kx
with k = 15 N/m. What will be its initial acceleration if it
is released from a point 20 cm away from the origin
(1) 5 m/s2
(2) 10 m/s2
2
(3) 3 m/s
(4) 15 m/s2
Q.8
In a rocket of mass 1000 kg fuel is consumed at a rate
of 40 kg/s. The velocity of the gases ejected from the
rocket is 5 × 104 m/s. The thrust on the rocket is
(1) 2 × 103 N
(2) 5 × 104 N
(3) 2 × 106 N
(4) 2 × 109 N
Q.9
Two balls of masses m1 and m2 are separated from each
other by a powder charge placed between them. The
whole system is at rest on the ground. Suddenly the
powder charge explodes and masses are pushed apart.
The mass m1 travels a distance s1 and stops. If the
coefficients of friction between the balls and ground
are same, the mass m2 stops after travelling the
distance
m1
(1) s 2 = m s1
2
(3) s 2 =
Q.10
m12
s1
m 22
m2
(2) s 2 = m s1
1
(4) s 2 =
m 22
s1
m12
A force vector applied on a mass is represented as

F = 6 ˆi - 8ˆj + 10kˆ and accelerates with 1m/s2 . What
will be the mass of the body
(1) 10 2 kg
(2) 2 10 kg
(3) 10 kg
(4) 20 kg
25
JEE - PHYSICS
Q.11
Q.12
Q.13
A cricket ball of mass 250 g collides with a bat with
velocity 10 m/s and returns with the same velocity
within 0.01 second. The force acted on bat is
(1) 25 N
(2) 50 N
(3) 250 N
(4) 500 N
A body of mass 2 kg is moving with a velocity 8 m/s on
a smooth surface. If it is to be brought to rest in
4 seconds, then the force to be applied is
(1) 8 N
(2) 4 N
(3) 2 N
(4) 1 N
The adjacent figure is the part of a horizontally
stretched net. section AB is stretched with a force of
10 N. The tensions in the sections BC and BF are
Q.18
A man is at rest in the middle of a pond on perfectly
smooth ice. He can get himself to the shore by making
use of Newton’s
(1) First law
(2) Second law
(3) Third law
(4) All the laws
Q.19
If a force of 250 N act on body, the momentum acquired
is 125 kg-m/s. What is the period for which force acts
on the body
(1) 0.5 sec
(2) 0.2 sec
(3) 0.4 sec
(4) 0.25 sec
Q.20
An aircraft is moving with a velocity of 300 ms–1. If all
the forces acting on it are balanced, then
(1) It still moves with the same velocity
(2) It will be just floating at the same point in space
(3) It will fall down instantaneously
(4) It will lose its velocity gradually
E
150°
150°
D
G
120°
C
F
B
120°
H
120°
A
(1) 10 N, 11 N
(2) 10 N, 6 N
(3) 10 N, 10 N
(4) Can’t calculate due to insufficient data
Q.14
Q.15
When the speed of a moving body is doubled
(1) Its acceleration is doubled
(2) Its momentum is doubled
(3) Its kinetic energy is doubled
(4) Its potential energy is doubled
2mν
t
(3) mt
Q.17
26
Q.22
A ball of mass m moves with speed v and it strikes
normally with a wall and reflected back normally, if its
time of contact with wall is t then find force exerted by
ball on wall
(1)
Q.16
Application of Force/Impulse, Statics and Dynamics
involving single system
Q.21
Two forces of magnitude F have a resultant of the
same magnitude F. The angle between the two forces
is
(1) 45°
(2) 120°
(3) 150°
(4) 60°
(2)
mν
t
mν
(4)
2t
A particle moves in the xy-plane under the action of a
force F such that the components of its linear
momentum p at any time t are px = 2cos t, py = 2sin t.
The angle between F and p at time t is
(1) 90°
(2) 0°
(3) 180°
(4) 30°
Swimming is possible on account of
(1) First law of motion
(2) Second law of motion
(3) Third law of motion
(4) Newton’s law of gravitation
Two blocks are connected by a string as shown in the
diagram. The upper block is hung by another string. A
force F applied on the upper string produces an
acceleration of 2 m/s2 in the upward direction in both
the blocks. If T and T' be the tensions in the two parts
F
of the string, then
T
(1) T = 70.8 N and T= 47.2 N
2 kg
(2) T = 58.8 N and T= 47.2 N
(3) T = 70.8 N and T = 58.8 N
T'
(4) T = 70.8 N and T = 0
4 kg
Q.23
Consider the following statements about the blocks
shown in the diagram that are being pushed by a
constant force on a frictionless table
F
3 kg
2 kg
1 kg
A. All blocks move with the same acceleration
B. The net force on each block is the same Which of
these statements are/is correct
(1) A only
(3) Both A and B
(2) B only
(4) Neither A nor B
Newton’s Laws of Motion and Friction
JEE - PHYSICS
Q.24
A rope of length L is pulled by a constant force F.
What is the tension in the rope at a distance x from the
end where the force is applied
(1)
FL
x
(2)
FL
(3)
L-x
Q.25
FL - x
L
6 kg
Fx
(4)
L-x
Three equal weights A, B and C of mass 2 kg each are
hanging on a string passing over a fixed frictionless
pulley as shown in the figure The tension in the string
connecting weights B and C is
Q.29
10 kg
(1) 24.5 N
(2) 2.45 N
(3) 79 N
(4) 73.5 N
A body of weight 2kg is suspended as shown in the
figure. The tension T 1 in the horizontal string
(in kg wt) is
30°
T1
2 kg-wt
A
(1) Zero
(3) 3.3 N
Q.26
B
(1) 2
C
(3) 2 3
(2) 13 N
(4) 19.6 N
Two masses of 4 kg and 5 kg are connected by a string
passing through a frictionless pulley and are kept on a
frictionless table as shown in the figure. The
acceleration of 5 kg mass is
5kg
(1) 49 m/s
(3) 19.5 m/s2
2
Q.27
Q.28
(2) 5. 44 m / s
(4) 2.72 m/s2
m2
(1) m + m + m T
1
2
3
m3
(2) m + m + m T
1
2
3
m1 + m 2
(3) m + m + m T
1
2
3
m 2 + m3
(4) m + m + m T
1
2
3
A light string passes over a frictionless pulley. To one
of its ends a mass of 6 kg is attached. To its other end
a mass of 10 kg is attached. The tension in the thread
will be
Newton’s Laws of Motion and Friction
(4) 2
A bird is sitting in a large closed cage which is placed
on a spring balance. It records a weight of 25 N. The
bird (mass m = 0.5 kg) flies upward in the cage with an
acceleration of 2 m/s2. The spring balance will now
record a weight of
(1) 24 N
(3) 26 N
2
Three solids of masses m1, m2 and m3 are connected
with weightless string in succession and are placed
on a frictionless table. If the mass m3 is dragged with a
force T, the tension in the string between m2 and m3 is
3 2
Constraint relation, Dynamics of multi system Spring
Q.30
A man is standing at a spring platform. Reading of
spring balance is 60 kg wt. If man jumps outside
platform, then reading of spring balance
(1) First increases then decreases to zero
(2) Decreases
(3) Increases
(4) Remains same
Q.31
4kg
(2)
3
Q.32
(2) 25 N
(4) 27 N
The tension in the spring is
5N
(1) Zero
(3) 5 N
Q.33
5N
(2) 2.5 N
(4) 10 N
A block of mass 4 kg is suspended through two light
spring balances A and B. Then A and B will read
respectively
(1) 4 kg and zero kg
A
(2) Zero kg and 4 kg
B
(3) 4 kg and 4 kg
(4) 2 kg and 2 kg
4kg
27
JEE - PHYSICS
Non Inertial Reference frame and Pseudo force
Q.34
A person is standing in an elevator. In which situation
he finds his weight less than actual when
(1) The elevator moves upward with constant
acceleration
(2) The elevator moves downward with constant
acceleration.
(3) The elevator moves upward with uniform velocity
(4) The elevator moves downward with uniform
velocity
Q.35
A body of mass 4 kg weighs 4.8 kg when suspended in
a moving lift. The acceleration of the lift is
(1) 9.80 ms–2 downwards (2) 9.80 ms–2 upwards
(3) 1.96 downwards
(4) 1.96 upwards
Q.36
The mass of a lift is 500 kg. When it ascends with an
acceleration of 2 m/s2, the tension in the cable will be
[g = 10 m/s2]
(1) 6000 N
(2) 5000 N
(3) 4000 N
(4) 50 N
Q.37
If in a stationary lift, a man is standing with a bucket
full of water, having a hole at its bottom. The rate of
flow of water through this hole is R0. If the lift starts to
move up and down with same acceleration and then
that rates of flow of water are Ru and Rd, then
Q.38
(1) R0 > Ru > Rd
(2) Ru > R0 > Rd
(3) Rd > R0 > Ru
(4)Ru > Rd >R0
A plumb line is suspended from a ceiling of a car
moving with horizontal acceleration of a. What will be
the angle of inclination with vertical
(1) tan– 1(a/g)
(2) tan–1 (g/a)
(3) cos–1 (a/g)
(4) cos–1 (g/a)
Q.42
T
(1) Move over the surface with constant velocity
(2) Move having accelerated motion over the surface
(3) Not move
(4) First it will move with a constant velocity for some
time and then will have accelerated motion
Q.43
Q.40
Q.41
28
A force of 98 N is required to just start moving a body
of mass 100 kg over ice. The coefficient of static friction
is
(1) 0.6
(2) 0.4
(3) 0.2
(4) 0.1
Maximum value of static friction is called
(1) Limiting friction
(2) Rolling friction
(3) Normal reaction
(4) Coefficient of friction
Two masses A and B of 10 kg and 5 kg respectively are
connected with a string passing over a frictionless
pulley fixed at the corner of a table as shown. The
coefficient of static friction of A with table is 0.2. The
minimum mass of C that may be placed on A to prevent
it from moving is
C
A
B
(1) 15 kg
(2) 10 kg
(3) 5 kg
(4) 12 kg
Q.44
Which of the following statements is not true
(1) The coefficient of friction between two surfaces
increases as the surface in contact are made rough
(2) The force of friction acts in a direction opposite to
the applied force
(3) Rolling friction is greater than sliding friction
(4) The coefficient of friction between wood and wood
is less than 1
Q.45
The blocks A and B are arranged as shown in the figure.
The pulley is frictionless. The mass of A is 10 kg. The
coefficient of friction of A with the horizontal surface
is 0.20. The minimum mass of B to start the motion will
be
FRICTION
Static Friction
Q.39
The coefficient of friction  and the angle of friction 
are related as
(1) sin  = 
(2) cos  = 
(3) tan  = 
(4) tan  = 
In the figure shown, a block of weight 10 N resting on
a horizontal surface. The coefficient of static friction
between the block and the surface is s = 0.4 . A force
of 3.5 N will keep the block in uniform motion, once it
has been set in motion. A horizontal force of 3 N is
applied to the block, then the block will
A
B
(1) 2 kg
(2) 0.2 kg
(3) 5 kg
(4) 10 kg
Newton’s Laws of Motion and Friction
JEE - PHYSICS
Q.46
A block A with mass 100 kg is resting on another block
B of mass 200 kg. As shown in figure a horizontal rope
tied to a wall holds it. The coefficient of friction
between A and B is 0.2 while coefficient of friction
between B and the ground is 0.3. The minimum required
force F to start moving B will be
Q.52
If s ,k and r are coefficients of static friction, sliding
friction and rolling friction, then
(1) s < k < r
(2) k < r < s
(3) r < k < s
(4) r = k = s
Q.53
A block of mass M = 5 kg is resting on a rough
horizontal surface for which the coefficient of friction
is 0.2. When a force F = 40 N is applied, the acceleration
of the block will be (g = 10 m/s2)
A
B
F
F
(1) 900 N
(2) 100 N
30°
(3) 1100 N (4) 1200 N
M
Q.47
What is the maximum value of the force F such that the
block shown in the arrangement, does not move
F
60°
(1) 20 N
(3) 12 N

Q.54
m=3kg
(2) 10 N
(4) 15 N
Kinetic friction
Q.48
A car is moving along a straight horizontal road with a
speed v0. If the coefficient of friction between the tyres
and the road is , the shortest distance in which the
car can be stopped is
 02
(1)
2g
Q.49
Q.50
Q.51
0
(2)
g
 0 
(3)  
 g 
2
(4)
0

A body B lies on a smooth horizontal table and another
body A is placed on B. The coefficient of friction
between A and B is . What acceleration given to B
will cause slipping to occur between A and B
(1) g
(1) 5.73 m/sec2
(3) 3.17 m/sec2
(2) g/
(3) /g
(4) g
A 500 kg horse pulls a cart of mass 1500 kg along a
level road with an acceleration of 1 ms –2. If the
coefficient of sliding friction is 0.2, then the force
exerted by the horse on cart in forward direction is
(1) 3000 N
(2) 4000 N
(3) 5000 N
(4) 4500 N
A horizontal force of 129.4 N is applied on a 10 kg
block which rests on a horizontal surface. If the
coefficient of friction is 0.3, the acceleration should be
(1) 9.8 m/s2
(2) 10 m/s2
2
(3) 12.6 m/s
(4) 19.6 m/s2.
Newton’s Laws of Motion and Friction
(2) 8.0 m/sec2
(4) 10.0 m/sec2
The coefficient of friction between a body and the
surface of an inclined plane at 45° is 0.5. If g = 9.8 m/s2,
the acceleration of the body downwards in m/s2 is
(1)
4.9
2
(2) 4.9 2
(3)
Q.55
(4) 4.9
The upper half of an inclined plane of inclination  is
perfectly smooth while the lower half is rough. A body
starting from the rest at top comes back to rest at the
bottom if the coefficient of friction for the lower half is
given by
(1)  = sin 
(2)  = cot 
(3)  = 2 cos 
(4)  = 2 tan 
Double Block Problem
Q.56
A 40 kg slab rests on a frictionless floor as shown in
the figure. A 10 kg block rests on the top of the slab.
The static coefficient of friction between the block and
slab is 0.60 while the kinetic friction is 0.40. The 10 kg
block is acted upon by a horizontal force of 100 N. If g
= 9.8 m/s2, the resulting acceleration of the slab will be
100 N
10 kg
40 kg
(1) 0.98 m/s2
(3) 1.52 m/s2
A
B
(2) 1.47 m/s2
(4) 6.1 m/s2
29
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JEE MAINS
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MODULE - 02
Newton’s Laws of Motion and Friction
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JEE - PHYSICS
EXERCISE-II
Scan for Video Solution
Q.1
Q.2
Q.3
NEWTON’S LAWS OF MOTION
Q.5
Let E,G and N represents the magnitude of
electromagnetic, gravitational and nuclear forces
between two protons at a given separation (1 fermi
meter). Then
(1) N < E < G
(2) E > N > G
(3) G > N > E
(4) N > E > G
A man getting down a running bus, falls forward
because(1) due to inertia of rest, road is left behind and man
reaches forward
(2) due to inertia of motion upper part of body continues
to be in motion in forward direction while feet
come to rest as soon as they touch the road
(3) he leans forward as a matter of habbit
(4) of the combined effect of all the three factors
stated in (1), (2) and (3)
A constant force F is applied in horizontal direction as
shown. Contact force between M and m is N and
between m and M’ is N’ then
M'>M
F
M m M'
smooth
(1) N = N’
(3) N’ > N
(2) N > N’
(4) cannot be determined
Q.6
n which of the following cases the net force is not
zero ?
(1) A kite skillfully held stationary in the sky
(2) A ball freely falling from a height
(3) An aeroplane rising upwards at an angle of 45º with
the horizontal with a constant speed
(4) A cork floating on the surface of water
Q.7
Two persons are holding a rope of negligible weight
tightly at its ends so that it is horizontal. A 15 kg weight
is attached to the rope at the mid point which now no
longer remains horizontal. The minimum tension
required to completely straighten the rope is :
(1) 15 kg
Which figure represents the correct F.B.D. of rod of
mass m as shown in figure :
15
kg
2
(3) 5 kg
(4) nfinitely large (or not possible)
(2)
T
(1)
R
T
(2)
Q.8
R
mg
mg
n the system shown in the figure, the acceleration of
the 1 kg mass and the tension in the string connecting
between A and B is :
T
(3)
R
(4) None of these
mg
Q.4
B 3kg
Two forces of 6N and 3N are acting on the two blocks
of 2kg and 1kg kept on frictionless floor. What is the
force exerted on 2kg block by 1kg block ?
6N
2kg 1kg
3N
30
(2) 2N
(4) 5N
C
A 1kg
(1)
g
8g
downwards ,
4
7
(3)
g
6
g
downwards , g (4)
upwards , g
7
7
2
///////////////////////////////////////////////
(1) 1N
(3) 4N
3kg
(2)
g
g
upwards,
4
7
Newton’s Laws of Motion and Friction
JEE - PHYSICS
Q.9
A body of mass 5 kg is suspended by the strings
making angles 60º and 30º with the horizontal -
Q.13
A particle of mass 50 gram moves on a straight line.
The variation of speed with time is shown in figure.
find the force acting on the particle at t = 2, 4 and 6
seconds.
v(m/s)
Q.10
(a) T1 = 25 N
(b) T2 = 25 N
15
(c) T1 = 25 3 N
(1) a, b
(2) a, d
(d) T2 = 25 3 N
(3) c, d
(4) b, c
10
5
0
The 50 kg homogeneous smooth sphere rests on the
30° incline A and bears against the smooth vertical
wall B. Calculate the contact forces at A and B.
A
B
Q.14
(4) NA =
Q.11
3
100
3
N , NB =
N , NB =
1000
3
3
500
3
500
50
N
3
(1)
Find out the reading of the weighing machine in the
following cases.
30º
(1) 10 3
(2) 10 2
(3) 20 3
(4) 30 3
Three blocks A, B and C are suspended as shown in
the figure. Mass of each blocks A and C is m. If system
is in equilibrium and mass of B is M, then :
 fixed 
M 2 g  sinβ 
(2)
M1 + M 2
 M 2 sinα – M 1 sinβ 
g
(3) 
M1 + M 2


Q.15
2
W kg
M
g
2k
M
W
30º
Q.12
N
N
3
N , NB =
N
1
1000
500
8 t(s)
M2
(3) NA =
3
N , NA =
6
Two masses M1 and M2 are attached to the ends of a
light string which passes over a massless pulley
attached to the top of a double inclined smooth plane
of angles of inclination  and . If M2 > M1 and  > 
then the acceleration of block M2 down the inclined
will be :
M
(2) NA =
1000
4
(1) 0.25 N along motion, zero, 0.25 opposite to motion
(2) 0.25 N along motion, zero, 0.25 along motion
(3) 0.25 N opposite to motion, zero, 0.25 along motion
(4) 0.25 N opposite to motion, zero, 0.25 opposite motion
30°
(1) NB =
2
M 1g  sinα 
M1 + M 2
(4) zero
The pulley arrangements shown in figure are identical,
the mass of the rope being negligible. In case I, the
mass m is lifted by attaching a mass 2m to the other
end of the rope. In case II, the mass m is lifted by
pulling the other end of the rope with a constant
downward force F = 2mg, where g is acceleration due
to gravity. The acceleration of mass in case I is :
///////////////////////////////////
F=2mg
m 2m
(I)
A
B
C
(1) M = 2m (2) M < 2 m (3) M > 2m (4) M = m
Newton’s Laws of Motion and Friction
m
(II)
(1) zero
(2) more than that in case II
(3) less than that in case II
(4) equal to that in case II
31
JEE - PHYSICS
Q.16
A fireman wants to slide down a rope. The rope can
Q.21
3
th of the weight of the man. With
4
what minimum acceleration should the fireman slide
down :
bear a tension of
(1)
Q.17
g
3
(2)
g
6
(3)
g
4
(4)
g
2
A rope of mass 5 kg is moving vertically in vertical
position with an upwards force of 100 N acting at the
upper end and a downwards force of 70 N acting at the
lower end. The tension at midpoint of the rope is
(1) 100 N
(2) 85 N
(3) 75 N
(4) 105 N
Q.18
A body is moving with a speed of 1 m/s and a force F
is needed to stop it in a distance x. If the speed of the
body is 3 m/s the force needed to stop it in the same
distance x will be
(1) 1.5 F
(2) 3F
(3) 6 F
(4) 9F
Q.19
Two blocks, each having mass M, rest on frictionless
surfaces as shown in the figure. If the pulleys are light
and frictionless, and M on the incline is allowed to
move down, then the tension in the string will be :
Two objects A and B of masses mA and mB are
attached by strings as shown in fig. If they are given
upward acceleration, then the ratio of tension T 1 : T 2
is (1) (mA + mB)/mB
(2) (mA + mB)/mA
mA + mB
(3) m - m
A
B
mA - mB
(4) m + m
A
B
Q.22
A monkey of mass 20 kg is holding a vertical rope. The
rope can break when a mass of 25 kg is suspended
from it. What is the maximum acceleration with which
the monkey can climb up along the rope?
(1) 7 ms–2
(2) 10 ms–2
(3) 5 ms–2
(4) 2.5 ms–2
Q.23
A body of mass 8 kg is hanging from another body of
mass 12 kg. The combination is being pulled by a string
with an acceleration of 2.2 m s–2. The tension T1 and T2
will be respectively:
(use g = 9.8m/s2 )
M

T1
M
2
(1) Mg sin 
3
3
(2) Mg sin 
2
Mgsinθ
2
(4) 2 Mg sin 
(3)
Q.20
fixed
12kg
8kg
(1) 200 N, 80 N
(3) 240 N, 96 N
Two masses are hanging vertically over frictionless
pulley. The acceleration of the two masses isQ.24
32
m1
(1) m g
2
m2
(2) m g
1
 m 2 - m1 
(3)  m + m  g
 1
2 
 m1 + m 2 
(4)  m - m  g
 2
1 
a
T2
(2) 220 N, 90 N
(4) 260 N, 96 N
At a given instant, A is moving with velocity of 5 m/s
upwards. What is velocity of B at the time
A
B
(1) 15 m/s
(3) 5 m/s 
(2) 15 m/s
(4) 5 m/s 
Newton’s Laws of Motion and Friction
JEE - PHYSICS
Q.25
Find the velocity of the hanging block if the velocities
of the free ends of the rope are as indicated in the
figure.
Q.29
If the tension in the cable supporting an elevator is
equal to the weight of the elevator, the elevator may
be (a) going up with increasing speed
(b) going down with increasing speed
(c) going up with uniform speed
(d) going down with uniform speed
(1) a, d
(2) a, b, c (3) c, d
(4) a, b
Q.30
n the given figure, what is the reading of the spring
balance ?
2m/s
1m/s
(1) 3/2 m/s 
(3) 1/2 m/s 
Q.26
(2) 3/2 m/s 
(4) 1/2 m/s 
In the arrangement shown in fig. the ends P and Q of
an unstretchable string move downwards with uniform
speed U. Pulleys A and B are fixed. Mass M moves
upwards with a speed.
1kg
1kg
(1) 10 N
B
A


Q.31
Q
P
M
(1) 2 U cos 
2U
(3)
cosθ
Q.27
(2) 20 N
(4) zero
Two blocks of masses M1 and M2 are connected to
each other through a light spring as shown in figure. If
we push mass M1 with force F and cause acceleration
a1 in right direction in mass M 1, what will be the
magnitude of acceleration in M2?
F
(2) U cos 
U
(4)
cosθ
(3) 5 N
M1
M2
///////////////////////////////////////////////////////////////
(1) F/M2
(3) a1
In the figure shown the velocity of lift is 2 m/s while
string is winding on the motor shaft with velocity
2 m/s and block A is moving downwards with a velocity
of 2 m/s, then find out the velocity of block B.
Q.32
(2) F/(M1 + M2)
(4) (F – M1a1)/M2
A massless spring balance is attached to 2 kg trolley
and is used to pull the trolley along a flat surface as
shown in the fig. The reading on the spring balance
remains at 10 kg during the motion. The acceleration
of the trolley is (Use g = 9.8 ms–2)
2kg
•
(1) 2 m/s 
(3) 4 m/s 
Q.28
(1) 4.9 ms–2
(3) 49 ms–2
(2) 2 m/s 
(4) 8 m/s 
Q.33
In the figure at the free end a force F is applied to keep
the suspended mass of 18 kg at rest. The value of F is-
•
Three masses of 1 kg , 6 kg and 3 kg are connected to
each other with threads and are placed on table as
shown in figure. What is the acceleration with which
the system is moving? Take g = 10m s–2 .
T1
T1
1kg
(1) 180 N
(2) 90 N
(3) 60 N
Newton’s Laws of Motion and Friction
(4) 30 N
(2) 9.8 ms–2
(4) 98 ms–2
(1) Zero
(3) 2 m s–2
6kg
T2
T2
3kg
(2) 1 m s–2
(4) 3 m s–2
33
JEE - PHYSICS
Q.34
In the figure a smooth pulley of negligible weight is
suspended by a spring balance. Weights of 1kg and
5 kg are attached to the opposite ends of a string
passing over the pulley and move with acceleration
because of gravity. During the motion, the spring
balance reads a weight of -
Q.39
A box 'A' is lying on the horizontal floor of the
compartment of a train running along horizontal rails
from left to right. At time 't', it decelerates. Then the
reaction R by the floor on the box is given best by
(1)
floor
R
(2)
(1) 6 kg
(2) less than 6 kg
(3) more than 6 kg
(4) may be more or less than 6 kg
Q.35
2 kg
A
floor
R
A
(3)
floor
R
Find the acceleration of 3 kg mass when acceleration
of 2 kg mass is 2 ms–2 as shown in figure.
3 kg
R
A
A
(4)
10N
floor
2ms
–2
(1) 3 ms
–2
Q.36
Q.37
Q.38
34
(2) 2 ms
–2
(3) 0.5 ms–2 (4) zero
Q.40
The ratio of the weight of a man in a stationary lift &
when it is moving downward with uniform acceleration
'a' is 3 : 2 . The value of 'a' is : (g = acceleration. due to
gravity)
(1) (3/2) g (2) g
(3) (2/3) g (4) g/3
FRICTION
A block is placed on a rough floor and a horizontal
force F is applied on it. The force of friction f by the
floor on the block is measured for different values of
F and a graph is plotted between them
(a) The graph is a straight line of slope 45°
(b) The graph is straight line parallel to the F axis
(c) The graph is a straight line of slope 45º for small
F and a straight line parallel to the F-axis for largeF.
(d) There is small kink on the graph
(1) c, d
(2) a, d
(3) a, b
(4) a, c
A monkey of mass m is climbing a rope hanging from
the roof with acceleration a. The coefficient of static
friction between the body of the monkey and the rope
is . Find the direction and value of friction force on
the monkey.
(1) Upward, F = m(g + a)
(2) downward, F = m(g + a)
(3) Upward, F = mg
(4) downward, F = mg
A body is placed on a rough inclined plane of
inclination . As the angle  is increased from 0º to 90º
the contact force between the block and the plane
(1) remains constant
(2) first remains constant then decreases
(3) first decreases then increases
(4) first increases then decreases
Q.41
Q.42
A chain is lying on a rough table with a fraction 1/n of
its length hanging down from the edge of the table. if
it is just on the point of sliding down from the table,
then the coefficient of friction between the table and
the chain is (1)
1
n
(2)
1
(n -1)
(3)
1
(n +1)
(4)
n -1
(n +1)
For the equilibrium of a body on an inclined plane of
inclination 45º. The coefficient of static friction will be
(1) greater than one
(2) less than one
(3) zero
(4) less than zero
Newton’s Laws of Motion and Friction
JEE - PHYSICS
Q.43
A block of mass 5 kg and surface area 2 m2 just begins
to slide down an inclined plane when the angle of
inclination is 30°. Keeping mass same, the surface area
of the block is doubled. The angle at which this starts
sliding down is :
(1) 30°
(3) 15°
Q.44
Q.45
(2) 60°
(4) none
A block moves down a smooth inclined plane of
inclination . Its velocity on reaching the bottom is v.
If it slides down a rough inclined plane of some
inclination, its velocity on reaching the bottom is
v/n, where n is a number greater than 0. The
coefficient of friction is given by -
1 

(1)  = tan   1 - 2 
 n 
A wooden block of mass m resting on a rough horizontal
table (coefficient of friction = ) is pulled by a force F
as shown in figure. The acceleration of the block
moving horizontally is :
F
1 

(2) = cot   1 - 2 
 n 
1 

(3)  = tan   1 - 2 
n


m
(1)
Fcosθ
m
(3)
F
(cosθ + μsinθ) – μg (4) none of these
m
(2)
1/2
1 

(4) = cot   1 - 2 
n


μFsinθ
M
1/2
EXERCISE-III
Scan for Video Solution
MCQ/COMPREHENSION/MATCHING/NUMERICAL
Q.1
A particle is resting on a smooth horizontal floor. At
t = 0 , a horizontal force starts acting on it. Magnitude
of the force increases with time according to law
F = . t , where  is a constant. For the figure shown
which of the following statements is/are correct ?
(A) tension T in the string increases with increase in 
(B) tension T in the string decreases with increase in 
(C) tension T > F if  > /3
(D) tension T > F if  > /4
Q.3
Two blocks A and B of equal mass m are connected
through a massless string and arranged as shown in
figure. Friction is absent everywhere. When the system
is released from rest.
A
(A) Curve 1 shows acceleration against time
(B) Curve 2 shows velocity against time
(C) Curve 2 shows velocity against acceleration
(D) none of these
Q.2
A string is wrapped round a log of wood and it is
pulled with a constant force F as shown in the figure.
Newton’s Laws of Motion and Friction
fixed
(A) tension in string is
30°
B
mg
2
mg
4
(C) acceleation of A is g/2
3
(D) acceleration of A is g
4
(B) tension in string is
35
ABOUT PHYSICS WALLAH
Alakh Pandey is one of
the most renowned
faculty in NEET & JEE
domain’s Physics. On his
YouTube channel, Physics
Wallah, he teaches the
Science courses of 11th and
12th standard to the
students aiming to appear
for the engineering and
medical entrance exams.
PW
Alakh
Pandey
Scan the QR Code
to download our app
PHYSICS WALLAH
- www.physicswallahalakhpandey.com | -www.physicswallah.live
www.physicswallah.live
- Physics Wallah |
- Physics Wallah - Alakh Pandey
PHYSICS
JEE MAINS
Dropper
& ADVANCED
MODULE - 02
Newton’s Laws of Motion and Friction
Video Solution will
be provided soon
Get complete class
11th JEE study
material(hard copies)
delivered at your home
at the lowest cost of
Rs 2800/- only
Order from book section of pw app
Physics Wallah
JEE - PHYSICS
Q.43
A block of mass 5 kg and surface area 2 m2 just begins
to slide down an inclined plane when the angle of
inclination is 30°. Keeping mass same, the surface area
of the block is doubled. The angle at which this starts
sliding down is :
(1) 30°
(3) 15°
Q.44
Q.45
(2) 60°
(4) none
A block moves down a smooth inclined plane of
inclination . Its velocity on reaching the bottom is v.
If it slides down a rough inclined plane of some
inclination, its velocity on reaching the bottom is
v/n, where n is a number greater than 0. The
coefficient of friction is given by -
1 

(1)  = tan   1 - 2 
 n 
A wooden block of mass m resting on a rough horizontal
table (coefficient of friction = ) is pulled by a force F
as shown in figure. The acceleration of the block
moving horizontally is :
F
1 

(2) = cot   1 - 2 
 n 
1 

(3)  = tan   1 - 2 
n


m
(1)
Fcosθ
m
(3)
F
(cosθ + μsinθ) – μg (4) none of these
m
(2)
1/2
1 

(4) = cot   1 - 2 
n


μFsinθ
M
1/2
EXERCISE-III
Scan for Video Solution
MCQ/COMPREHENSION/MATCHING/NUMERICAL
Q.1
A particle is resting on a smooth horizontal floor. At
t = 0 , a horizontal force starts acting on it. Magnitude
of the force increases with time according to law
F = . t , where  is a constant. For the figure shown
which of the following statements is/are correct ?
(A) tension T in the string increases with increase in 
(B) tension T in the string decreases with increase in 
(C) tension T > F if  > /3
(D) tension T > F if  > /4
Q.3
Two blocks A and B of equal mass m are connected
through a massless string and arranged as shown in
figure. Friction is absent everywhere. When the system
is released from rest.
A
(A) Curve 1 shows acceleration against time
(B) Curve 2 shows velocity against time
(C) Curve 2 shows velocity against acceleration
(D) none of these
Q.2
A string is wrapped round a log of wood and it is
pulled with a constant force F as shown in the figure.
Newton’s Laws of Motion and Friction
fixed
(A) tension in string is
30°
B
mg
2
mg
4
(C) acceleation of A is g/2
3
(D) acceleration of A is g
4
(B) tension in string is
35
JEE - PHYSICS
Q.4
Two men of unequal masses hold on to the two sections
Q.6
of a light rope passing over a smooth light pulley.
Which of the following are possible?
In the system shown in the figure m1 > m2 . System is
held at rest by thread BC . Just after the thread BC is
burnt :
/////////////////////
spring
k
m1 A
(B) The heavier man is stationary while the lighter man
m  m2 
 g
equal to  1
 m1  m2 
climbs with some acceleration
(C) The two men slide with the same acceleration in
(C) acceleration of m1 will be equal to zero
the same direction
(D) magnitude of acceleration of two blocks will be
(D) The two men move with accelerations of the same
nonzero and unequal.
magnitude in opposite directions
Q.7
Two blocks of masses 10 kg and 20 kg are connected
by a light spring as shown. A force of 200 N acts on
connected by an ideal string as shown in the figure.
the 20 kg mass as shown. At a certain instant the
Neglect the masses of the pulleys and effect of friction.
acceleration of 10 kg mass is 12 ms–2 towards right
(g = 10 m/s2)
direction.
(A) At that instant the 20 kg mass has an acceleration
B
A
45°
C
(B) magnitude of acceleration of both blocks will be
man slides with some acceleration
Two blocks A and B of mass 10 kg and 40 kg are
////////////
(A) acceleration of m2 will be upwards
(A) The lighter man is stationary while the heavier
Q.5
B m2
Fixed
of 12 ms–2.
(B) At that instant the 20 kg mass has an acceleration
45°
of 4 ms–2 .
(C) The stretching force in the spring is 120 N.
(D) The collective system moves with a common
5
(A) The acceleration of block A is
(B) The acceleration of block B is
(C) The tension in the string is
(D) The tension in the string is
36
2
5
2 2
125
2
150
2
acceleration of 30 ms–2 when the extension in the
ms–2
ms–2
connecting spring is the maximum.
Q.8
A particle stays at rest as seen in a frame. We can
conclude that
(A) the frame is inertial.
N
(B) resultant force on the particle is zero.
(C) if the frame is inertial then the resultant force on
the particle is zero.
N
(D) if the frame is noninertial then there is a nonzero
resultant force.
Newton’s Laws of Motion and Friction
JEE - PHYSICS
Q.9
A man pulls a block heavier than himself with a light
horizontal rope. The coefficient of friction is the same
between the man and the ground, and between the
block and the ground.
(A) The block will not move unless the man also moves
(B) The man can move even when the block is
stationary
(C) If both move, the acceleration of the man is greater
than the acceleration of the block
(D) None of the above assertions is correct
Q.10
Car is accelerating with acceleration = 20 m/s2. A box
that is placed inside the car, of mass m = 10 kg is put in
contact with the vertical wall as shown. The friction
coefficient between the box and the wall is  = 0.6.
  0.6
10kg
Q.12
Mass of the object in kg and the normal force acting
on the block due to weighing machine are:
(A) 60 kg, 450 N
(B) 40 kg, 150 N
(C) 80 kg, 400 N
(D) 10 kg, zero
Q.13
If lift is stopped and equilibrium is reached. Reading
of weighing machine and spring balance will be :
(A) 40 kg, zero
(B) 10 kg, 20 kg
(C) 20 kg, 10 kg
(D) zero, 40 kg
Q.14
Find the acceleration of the lift such that the weighing
machine shows its true weight.
20m/s2
(A) The acceleration of the box will be 20 m/sec2
(B) The friction force acting on the box will be 100 N
(C) The contact force between the vertical wall and
the box will be 100 5 N
(D) The net contact force between the vertical wall
and the box is only of electromagnetic in nature.
Q.11
The value(s) of mass m for which the 100 kg block
remains is static equilibrium is
100
  0.3
m
Q.15
(B) 37 kg
(D) 85 kg
Comprehension # 1 (Q. No. 12 to 14 )
Figure shows a weighing machine kept in a lift. Lift is
moving upwards with acceleration of 5 m/s2. A block is
kept on the weighing machine. Upper surface of block
is attached with a spring balance. Reading shown by
weighing machine and spring balance is 15 kg and 45
kg respectively.
Answer the following questions. Asume that the
weighing machine can measure weight by having
negligible deformation due to block, while the spring
balance requires larger expansion: (take g = 10 m/s2)
Newton’s Laws of Motion and Friction
45
m/s2
4
(B)
85
m/s2
4
(C)
22
m/s2
4
(D)
60
m/s2
4
Comprehension # 2 (Q. No. 15 to 17)
A small block of mass 1 kg starts moving with constant
velocity 2 m/s on a smooth long plank of mass 10 kg
which is also pulled by a horizontal force F = 10 t N
where t is in seconds and F is in newtons. (the initial
velocity of the plank is zero).
37°
(A) 35 kg
(C) 83 kg
(A)
Displacement of 1 kg block with respect to plank at
the instant when both have same velocity is
4
(A) 4 m
(B) 4 m
3
(C)
Q.16
Q.17
8
m
3
(D) 2 m
The time (t  0) at which displacement of block and
plank with respect to ground is same will be :
(A) 12 s
(B) 2 3 s
(C) 3 3 s
(D)
3/2 s
Relative velocity of plank with respect to block when
acceleration of plank is 4 m/s2 will be (A) Zero
(B) 10 m/s
(C) 6 m/s
(D) 8 m/s
37
JEE - PHYSICS
Comprehension # 3 (Q. No. 18 to 20 )
An object of mass 2 kg is placed at rest in a frame (S1)
moving with velocity 10iˆ  5 ˆj m/s and having
acceleration 5iˆ  10 ˆjm / s 2 . The object is also seen
Comprehension # 4 (Q. No. 21 and 22)
Imagine a situation in which the horizontal surface of
block M0 is smooth and its vertical surface is rough
with a coefficient of friction 
Smooth M
by an observer standing in a frame (S2) moving with
velocity 5iˆ  10 ˆjm / s
Q.18
(A) F = – 10 î – 20 ˆj due to acceleration of frame S1
Q.21
Identify the correct statement(s)
(A) If F=0, the blocks cannot remain stationary
(B) For one unique value of F, the blocks M and m
remain stationary with respect to M0
(C) The limiting friction between m and M 0 is
independent of F
(D) There exist a value of F at which friction force is
equal to zero.
Q.22
In above problem, choose the correct value(s) of F
which the blocks M and m remain stationary with
respect to M0
(C) F = – 10 i – 30 j due to acceleration of frame S1
(D) none of these
Q.20
Q.23
Calculate net force acting on object with respect to S2
frame.
(A) F = 20iˆ  20ˆj
(B) F = 10iˆ  20 ˆj
(C) F = 5 ˆi  2 0 ˆj
(D) F = 1 0 ˆi  5 ˆj
m
Rough
(B) F = – 20 î – 20 ˆj due to acceleration of frame S1
Q.19
Mo
F
Calculate 'Pseudo force' acting on object. Which frame
is responsible for this force.
Calculate net force acting on object with respect of S1
frame.
(A) 0
(B) 1
(C) 2
(D) none of these
(A) (M0  M  m)
g

(B)
(C) (M0  M  m)
mg
M
(D) none of these
m(M 0  M  m)g
M – m
Column- I gives four different situation. In final statement of each situation two vector quantities are compared. The
result of comparison is given in column-II. Match the statement in column-I with the correct comparison(s) in column-II:
Column 
Column 
(A) Stone is projected from ground at an angle  with horizontal (  90°).
(p) same in magnitude
Neglect the effect of air friction. Then between the two instants
when it is at same height (above ground), its average velocity and
horizontal component of velocity are
(B) For four particles A, B, C & D, the velocities of one with respect to other
(q) different in magnitude


are given as V DC is 20 m/s towards north, VBC is 20 m/s towards east and



V BA is 20 m/s towards south. Then VBC and V AD are
(C) Two blocks of masses 4 and 8 kg are placed on ground as shown
(r) same in direction
. Then the net force exerted by earth on block
of mass 8 kg and normal reaction exerted by 8 kg block on
earth are (note that earth
includes ground)
(D) For a particle undergoing rectilinear motion with uniform acceleration, the
magnitude of displacement is half the distance covered in some time in
terval. The magnitude of final velocity is less than magnitude of initial
velocity for this time interval. Then the initial velocity and average velocity
for this time interval are
38
(s) opposite in direction
Newton’s Laws of Motion and Friction
JEE - PHYSICS
Q.24
Column-I gives four different situations involving two blocks of mass m1 and m2 placed in different ways on a smooth
horizontal surface as shown. In each of the situations horizontal forces F1 and F2 are applied on blocks of mass m1 and m2
respectively and also m2 F1 < m1 F2. Match the statements in column I with corresponding results in column-II.
Column 
Column 
(A)
. Both the blocks
m1 m 2  F1
F2 
(p) m  m  m  m 
1
2 
1
2 
are connected by massless inelastic string. The
magnitude of tension in the string is
. Both the blocks
(B)
m1 m 2  F1
F2 
(q) m  m  m  m 
1
2 
1
2 
are connected by massless inelastic string. The
magnitude of tension in the string is
(C)
. The magnitude
m1 m 2
(r) m  m
1
2
 F2
F 
 1 

 m 2 m1 
of normal reaction between the blocks is
. The magnitude
(D)
of normal reaction between the blocks is
 F1  F2 
(s) m 1 m 2  m  m 
 1
2 
Q.27
NUMERICAL VALUE BASED
Q.25
The mass of the body which is hanging on the rope
attached to the movable pulley is four times as much as
the mass of the body which is fixed to the ground. At a
given instant the fixed body is released. What is its
initial acceleration (in m/s2) ? (The mass of the pulleys
and the ropes is negligible.)
In ideal spring, with a pointer attached to its end, hangs
next to a scale. With a 100 N weight attached and at
rest, the pointer indicates '40' on the scale as shown.
Using a 200 N weight instead results in 60 on the scale.
Using an unknown weight 'X' instead results in '30' on
the scale. Find value of 'X' (in newton).
0
100N
4m
Q.26
m
40
Q.28
A block of mass 5kg is being pulled up by a constant
force of 90 N starting at ground from rest. 2 sec later, a
part of the block having mass 3kg falls off. Find the
distance between the two parts when 3kg part hits the
ground.
Q.29
If MB = 10 kg, find the value of MA (in kg), so that block
A remains in equilibrium. The pulleys are ideal.
If mass M is 2 kg what is the tension in string AC ? (in N)
A
30°
B
60°
C
A
M
Newton’s Laws of Motion and Friction
B
39
JEE - PHYSICS
Q.30
The scale in Figure is being pulled on by three ropes.
Q.32
Compute the least acceleration (in m/s2) with which a
50 kg man can slide down a rope, if the rope can withstand a tension of 300 N.
Q.33
The conveyor belt is moving at speed 4 m/s. If the
coefficient of static friction between the conveyor and
the package B of mass 10 kg is 0.2, determine the
shortest time (in sec) the belt can stop so that the
package does not slide on the belt.
What net force (in N) does the spring scale read?
37°
37°
75N
B
75N
150N
Q.31
A rope has a length of 12 m and a mass of 16 kg. The
Q.34
rope hangs from a rigid support. An operator whose
A force P pulls a block at a constant speed across the
floor. What is the magnitude of the force (in N)?
mass is 80 kg slides down the rope at a constant speed
37°
of 0.8 m/s. What is the tension (in N) in the rope at a
60kg
point 6 m from the top when the man has slid below this
µ = 1/3
point?
EXERCISE-IV
Scan for Video Solution
JEE-MAIN
PREVIOUS YEAR’S
Q.1
Two masses m1 = 5kg and m2 = 10kg, connected by an
inextensible string over a frictionless pulley, are moving
as shown in the figure. The coefficient of friction of
horizontal surface is 0.15. The minimum weight m that
should be put on top of m2 to stop the motion is :
[JEE Main-2018]
Q.3
(1) 200 N
(2) 140 N
(3) 70 N
(4) 100 N
block of mass 10 kg is kept on a rough inclined plane
as shown in the figure. A force of 3 N is applied on the
block. The coefficient of static friction between the
plane and the block is 0.6. What should be the minimum
value of force P, such that the block does not move
downward ? (take = g=10ms–2)
[JEE Main - 2019 (January)]
P
(1) 27.3 kg
(3) 10.3 kg
Q.2
40
(2) 43.3 kg
(4) 18.3 kg
A mass of 10 kg is suspended vertically by a rope from
the roof. When a horizontal force is applied on the rope
at some point, the rope deviated at an 45° at the roof
point. If the suspended mass is at equilibrium, the
magnitude of the force applied is (g = 10 ms -2)
[JEE Main - 2019 (January)]
3N 45º
(1) 32 N
(2) 18 N
(3) 23 N
(4) 25 N
Newton’s Laws of Motion and Friction
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MODULE - 02
Newton’s Laws of Motion and Friction
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JEE - PHYSICS
Q.30
The scale in Figure is being pulled on by three ropes.
Q.32
Compute the least acceleration (in m/s2) with which a
50 kg man can slide down a rope, if the rope can withstand a tension of 300 N.
Q.33
The conveyor belt is moving at speed 4 m/s. If the
coefficient of static friction between the conveyor and
the package B of mass 10 kg is 0.2, determine the
shortest time (in sec) the belt can stop so that the
package does not slide on the belt.
What net force (in N) does the spring scale read?
37°
37°
75N
B
75N
150N
Q.31
A rope has a length of 12 m and a mass of 16 kg. The
Q.34
rope hangs from a rigid support. An operator whose
A force P pulls a block at a constant speed across the
floor. What is the magnitude of the force (in N)?
mass is 80 kg slides down the rope at a constant speed
37°
of 0.8 m/s. What is the tension (in N) in the rope at a
60kg
point 6 m from the top when the man has slid below this
µ = 1/3
point?
EXERCISE-IV
Scan for Video Solution
JEE-MAIN
PREVIOUS YEAR’S
Q.1
Two masses m1 = 5kg and m2 = 10kg, connected by an
inextensible string over a frictionless pulley, are moving
as shown in the figure. The coefficient of friction of
horizontal surface is 0.15. The minimum weight m that
should be put on top of m2 to stop the motion is :
[JEE Main-2018]
Q.3
(1) 200 N
(2) 140 N
(3) 70 N
(4) 100 N
block of mass 10 kg is kept on a rough inclined plane
as shown in the figure. A force of 3 N is applied on the
block. The coefficient of static friction between the
plane and the block is 0.6. What should be the minimum
value of force P, such that the block does not move
downward ? (take = g=10ms–2)
[JEE Main - 2019 (January)]
P
(1) 27.3 kg
(3) 10.3 kg
Q.2
40
(2) 43.3 kg
(4) 18.3 kg
A mass of 10 kg is suspended vertically by a rope from
the roof. When a horizontal force is applied on the rope
at some point, the rope deviated at an 45° at the roof
point. If the suspended mass is at equilibrium, the
magnitude of the force applied is (g = 10 ms -2)
[JEE Main - 2019 (January)]
3N 45º
(1) 32 N
(2) 18 N
(3) 23 N
(4) 25 N
Newton’s Laws of Motion and Friction
JEE - PHYSICS
Q.4
A block kept on a rough inclined plane, as shown in the
Q.6
Two blocks A and B of masses mA = 1 kg and mB = 3 kg
figure, reamins at rest upto a maximum force 2 N down
are kept on the table as shown in figure. The coefficient
the inclined plane. The maximum external force up the
of friction between A and B is 0.2 and between B and
inclined plane that does not move the block is 10 N.
the surface of the table is also 0.2. The maximum force
The coefficient of static friction between the block and
F that can be applied on B horizontally, so that the
the plane is :
block A does not slide over the block B is:
[JEE Main - 2019 (January)]
(Take g = 10 m/s2 )
N
10
[JEE Main-2019(April)]
A
B
2N 30º
(1)
(3)
3
2
1
2
(2)
(4)
3
4
(1) 16 N
Q.7
(2) 40 N
(3) 12 N
(4) 8 N
A block of mass 5 kg is (i) pushed in case (A) and (ii)
pulled in case (B). by a force F = 20N. Making an angle
2
3
of 30º with the horizontal, as shown in the figures. The
coefficient of friction between the block and floor is
Q.5
A ball is thrown upward with an initial velocity V0 from
= 0.2. The difference between the accelerations of the
the surface of the earth. The motion of the ball is
block, in case (B) and case (A) will be :(g = 10 ms–2)
affected by a drag force equal to mv2 (where m is mass
[JEE Main-2019(April)]
of the ball, v is its instantaneous velocity and  is a
constant). Time taken by the ball to rise to its zenith is
(1)
(A)
  
1
sin 1 
V0 
g
 g 
  
1
tan 1 
V0 
(2)
g
 g 
(3)
F=20N
[JEE Main - 2019(April)]
:
 2 
1
tan 1 
V0 
2 g
 g 
30º
30º
(B)
F=20N
Q.8
(1) 0 ms–2
(2) 0.8 ms–2
(3) 0.4 ms–2
(4) 3.2 ms–2
A mass of 10 kg is suspended by a rope of length 4m,
from the ceiling. a force F is applied horizontally at the
mid-point of the rope such that the top half of the rope
makes an angle of 45o with the vertical. Then F equals:
(Take g = 10ms–2 and the rope to be massless)
(4)

1
 
ln 1 
V0
g 
g 
Newton’s Laws of Motion and Friction
[JEE Main-2020 (January)]
(1) 75 N
(2) 90 N
(3) 100 N
(4) 70 N
41
JEE - PHYSICS
Q.9
A small ball of mass m is thrown upward with velocity
Q.12
u from the ground. The ball experiences a resistive force
mkv2, where v is its speed. The maximum height attained
[JEE Main-2020 (September)]
by the ball is
(1)
1  ku 2
(2) K ln 1  2g

1
ku 2
tan 1
2K
g
 ku 2
1
ln
(3) 2K 1  g







A block kept on a rough inclined plane, as shown in the
figure, reamins at rest upto a maximum force 2 N down
the inclined plane. The maximum external force up the
inclined plane that does not move the block is 10 N.
The coefficient of static friction between the block and
the plane is :
[JEE Main - 2019 (January)]
N
10
2
1
1 ku
tan
(4)
K
2g
2N 30º
Q.10
Two masses m1 = 5kg and m2 = 10kg, connected by an
inextensible string over a frictionless pulley, are moving
(1)
as shown in the figure. The coefficient of friction of
horizontal surface is 0.15. The minimum weight m that
should be put on top of m2 to stop the motion is :
[JEE Main-2018]
Q.11
(1) 27.3 kg
(2) 43.3 kg
(3) 10.3 kg
(4) 18.3 kg
Q.13
(2)
3
4
(3)
1
2
(4)
2
3
Two blocks A and B of masses mA = 1 kg and mB = 3 kg
are kept on the table as shown in figure. The coefficient
of friction between A and B is 0.2 and between B and
the surface of the table is also 0.2. The maximum force
F that can be applied on B horizontally, so that the
block A does not slide over the block B is:
(Take g = 10 m/s2 )
[JEE Main-2019(April)]
A
B
A block of mass 10 kg is kept on a rough inclined plane
as shown in the figure. A force of 3 N is applied on the
(1) 16 N
(3) 12 N
block. The coefficient of static friction between the
plane and the block is 0.6. What should be the minimum
value of force P, such that the block does not move
downward ? (take = g=10ms )
–2
[JEE Main - 2019 (January)]
P
Q.14
(2) 40 N
(4) 8 N
A block of mass 5 kg is (i) pushed in case (A) and (ii)
pulled in case (B). by a force F = 20N. Making an angle
of 30º with the horizontal, as shown in the figures. The
coefficient of friction between the block and floor is
= 0.2. The difference between the accelerations of the
block, in case (B) and case (A) will be :(g = 10 ms–2)
[JEE Main-2019(April)]
F=20N
3N 45º
42
3
2
(A)
(1) 32 N
(2) 18 N
(3) 23 N
(4) 25 N
30º
(B)
30º
F=20N
(1) 0 ms–2
(2) 0.8 ms–2
(3) 0.4 ms–2 (4) 3.2 ms–2
Newton’s Laws of Motion and Friction
JEE - PHYSICS
Q.15
A block starts moving up an inclined plane of inclination
30° with an initial velocity of v . It comes back to its
initial position with velocity
Q.3
v0
. The value of the
2
coefficient of kinetic friction between the block and
the inclined plane is close to
to I is _________.
Q.16
1
. The nearest integer
1000
[JEE Main-2020 (September)]
F
An insect is at the bottom of a hemispherical ditch of
radius 1 m. It crawls up the ditch but starts slipping
after it is at height h from the bottom. If the coefficient
of friction between the ground and the insect is 0.75,
then h is (g = 10 ms–2)
[JEE Main-2020 (September)]
(1) 0.45 m
(2) 0.80 m
(3) 0.20 m
(4) 0.60 m
JEE-ADVANCED
PREVIOUS YEAR’S
Q.1
The pulleys and strings shown in the figure are
smooth and of negligible mass for the system to
remain in equilibrium, the angle  should be
[JEE-2002]
m
Q.2
(C)
Q.4
F
x
(B) 2m
2
a  x2
F x
2m a
(D)
a
(A) gk
(B) 2gk
2a
(B) 30º
(D) 60º
a
(D) 4gk
(C) gk
Q.5
What is the maximum value of the force F such that the
block shown in the arrangement, does not move :
F

60º
(A) g/2 upwards , g downwards
(B) g upwards, g/ 2 downwards
(C) g upwards , 2g downwards
(D) 2g upwards , g downwards
Newton’s Laws of Motion and Friction
F a2  x 2
2m
x
A piece of wire is bent in the shape of a parabola y = kx2
(y-axis vertical) with a bead of mass m on it. The bead
can slide on the wire without friction. It stays at the
lowest point of the parabola when the wire is at rest.
The wire is now accelerated parallel to the x-axis with a
constant acceleration a. The distance of the new
equilibrium position of the bead, where the bead can
stay at rest with respect to the wire, from the y-axis is
a
m
System shown in figure is in equilibrium and at rest.
The spring and string are massless Now the string
is cut. The acc7eleration of mass 2m and m just after
the string is cut will be : [JEE-2006]
a
F
a
(A) 2m
2
a  x2
2m
(A) 0º
(C) 45º
m
P
a

m
Two particles of mass m each are tied at the ends of a
light string of length 2a. The whole system is kept on a
frictionless horizontal surface with the string held tight
so that each mass is at a distance ‘a’ from the centre P
(as shown in the figure). Now, the mid-point of the
string is pulled vertically upwards with a small but
constant force F. As a result, the particles move towards
each other on the surface. The magnitude of
acceleration, when the separation between them
becomes 2x, is
[JEE- 2007]
m
1
2 3
3 kg
///////////////////////////////////
[ JEE - 2003]
(A) 20 N
(C) 12 N
(B) 10 N
(D) 15 N
43
JEE - PHYSICS
Q.6
Two blocks A and B of equal masses are sliding down
Q.9
A block of mass m is on inclined plane of angle . The
along straight parallel lines on an inclined plane of 45°.
coefficient of friction between the block and the plane
Their coefficients of kinetic friction are A = 0.2 and
is  and tan > . The block is held stationary by ap-
B = 0.3 respectively. At t = 0, both the blocks are at rest
plying a force P parallel to the plane. The direction of
and block A is
force pointing up the plane is taken to be positive. As
2 meter behind block B. The time and
P is varied from P1 = mg(sin – cos) to P2 = mg(sin
distance from the initial position where the front faces
+ cos), the frictional force f versus P graph will look
of the blocks come in line on the inclined plane as shown
A
like :
[JEE-2004]
in figure. (Use g = 10 ms–2.)
[JEE-2010]
2m
B
Fixed
45°
(A)
Q.7
(A) 2s, 8 2 m
(B) 2 s, 7m
(C) 2 s, 7 2 m
(D) 2s, 7/ 2 m
A disc is kept on a smooth horizontal plane with its
plane parallel to horizontal plane. A groove is made in
(B)
the disc as shown in the figure. The coefficient of friction between mass m and surface of the groove is 2/5
and sin  = 3/5. Find the acceleration of mass with
respect to the frame of reference of the disc.
[JEE-2006]
(C)
ao = 25 m/s2

Q.8
Statement-1 It is easier to pull a heavy object than to
push it on a level ground.
[JEE-2008]
(D)
Statement-2 The magnitude of frictional force depends
on the nature of the two surfaces in contact.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement -2 is False
(D) Statement-1 is False, Statement-2 is True.
44
Q.10
A block is moving on an inclined plane making an angle
45° with the horizontal and the coefficient of friction is
. The force required to just push it up the inclined
plane is 3 times the force required to just prevent it
from sliding down. If we define N = 10 , then N is
[JEE-2011]
Newton’s Laws of Motion and Friction
JEE - PHYSICS
Q.11
A small block of mass of 0.1 kg lies on a fixed inclined
plane PQ which makes an angle  with the horizontal. A
horizontal force of 1 N on the block through its center
of mass as shown in the figure. The block remains
stationary if (take g = 10 m/s2)
(A) = 45°
[IIT-JEE-2012]
(B)  > 45° and a frictional force acts on the block
towards P.
(C)  > 45° and a frictional force acts on the block
towards Q.
(D)  < 45° and a frictional force acts on the block
towards Q.
Q.12
A block of mass m1 = 1 kg another mass m2 = 2 kg , are
placed together (see figure) on an inclined plane with
angle of inclination . Various values of  are given in
List I. The coefficient of friction between the block m1
and the plane is always zero. The coefficient of static
and dynamic friction between the block m2 and the plane
are equal to  = 0.3. In List II expression for the friction
on block m2 given. Match the correct expression of the
friction in List II with the angles given in List I, and
choose the correct option. The acceleration due to
gravity is denoted by g
[JEE Advanced-2014]
[Useful information : tan(5.5°)  0.1 ; tan (11.5°)  0.2 ;
tan(16.5º  0.3)]
Q.13
A particle of mass m is initially at rest at the origin. It is
subjected to a force and starts moving along the x-axis.
Its kinetic energy K changes with time as dK/dt = t,
where  is a positive constant of appropriate
dimensions. Which of the following statements is (are)
true ?
[JEE Advanced-2018]
(A) The force applied on the particle is constante
(B) The speed of the particle is proportional to ime
(C) The distance of the particle from the origin increases
linerarly with time
(D) The force is conservative
Q.14
A block of mass 2M is attached to a massless spring
with spring-constant k. This block is connected to two
other blocks of masses M and 2M using two massless
pulleys and strings. The accelerations of the blocks
are a1, a2 and a3 as shown in figure. The system is released from rest with the spring in its unstretched state.
The maximum extension of the spring is x0. Which of
the following option(s) is/are correct ?
[g is the acceleration due to gravity. Neglect friction]
[JEE Advanced 2019]
2M
a1
a2
M
(A) x0 =
a3
2M
4Mg
k
(B) When spring achieves an extension of
x0
for the
2
first time, the speed of the block connected to the spring
is 3g
List-I
 = 5°
 = 10°
 = 15°
 = 20°
P.
Q.
R.
S.
Code :
(A) P-1, Q-1, R-1,S-3
(C) P-2, Q-2, R-2,S-4
List-II
1. m2g sin 
2. (m1 + m2)g sin 
3. m2g cos 
4. (m1 + m2)g cos 
(B) P-2, Q-2, R-2,S-3
(D) P-2, Q-2, R-3,S-3
Newton’s Laws of Motion and Friction
M
5k
(C) a2 – a1 = a1 – a3
(D) At an extension of
x0
of the spring, the magnitude
4
of acceleration of the block connected to the spring is
3g
10
45
JEE - PHYSICS
Q.15
A block of weight 100 N is suspended by copper and
steel wires of same cross sectional area 0.5 cm2 and,
length
3 m and 1 m, respectively. Their other ends
are fixed on a ceiling as shown in figure. The angles
subtended by copper and steel wires with ceiling are
30° and 60°, respectively. If elongation in copper wire
is (C) and elongation in steel wire is (S), then the
 C
ratio  is ______.
S
46
[Young’s modulus for copper and steel are
1 × 1011 N/m2 and 2 × 1011 N/m2 respectively]
60º
Steel wire
1m
3m
30º
Copper wire
Block
[JEE Advanced 2019]
Newton’s Laws of Motion and Friction
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JEE - PHYSICS
ANSWER KEY
EXERCISE-I
Q.1 (3)
Q.11 (4)
Q.21 (2)
Q.31 (3)
Q.41 (1)
Q.51 (2)
Q.2 (3)
Q.12 (2)
Q.22 (1)
Q.32 (3)
Q.42 (3)
Q.52 (3)
Q.3 (2)
Q.13 (3)
Q.23 (1)
Q.33 (3)
Q.43 (1)
Q.53 (1)
Q.4 (2)
Q.14 (2)
Q.24 (2)
Q.34 (2)
Q.44 (3)
Q.54 (1)
Q.5 (3)
Q.15 (1)
Q.25 (2)
Q.35 (4)
Q.45 (1)
Q.55 (4)
Q.6 (2)
Q.16 (1)
Q.26 (2)
Q.36 (1)
Q.46 (3)
Q.56 (1)
Q.7 (2)
Q.17 (3)
Q.27 (3)
Q.37 (2)
Q.47 (1)
Q.8 (3)
Q.18 (3)
Q.28 (4)
Q.38 (1)
Q.48 (1)
Q.9 (3)
Q.19 (1)
Q.29 (3)
Q.39 (3)
Q.49 (1)
Q.10
Q.20
Q.30
Q.40
Q.50
(1)
(1)
(1)
(4)
(4)
Q.7 (4)
Q.17 (2)
Q.27 (4)
Q.37 (1)
Q.8 (3)
Q.18 (4)
Q.28 (2)
Q.38 (1)
Q.9 (2)
Q.19 (3)
Q.29 (3)
Q.39 (3)
Q.10
Q.20
Q.30
Q.40
(2)
(3)
(1)
(2)
Q.5 (A, B, D) Q.6 (A, C)
Q.7 (B,C)
Q.8 (C, D)
Q.14 (A)
Q.17 (C)
Q.18 (A)
EXERCISE-II
Q.1 (4)
Q.11 (1)
Q.21 (1)
Q.31 (4)
Q.41 (2)
Q.2 (2)
Q.12 (2)
Q.22 (4)
Q.32 (3)
Q.42 (1)
Q.3 (3)
Q.13 (1)
Q.23 (3)
Q.33 (3)
Q.43 (1)
Q.4 (3)
Q.14 (3)
Q.24 (1)
Q.34 (2)
Q.44 (3)
Q.5 (2)
Q.15 (3)
Q.25 (1)
Q.35 (2)
Q.45 (1)
Q.6 (2)
Q.16 (3)
Q.26 (4)
Q.36 (4)
EXERCISE-III
MCQ/COMPREHENSION/MATCHING/NUMERICAL
Q.1 (A, B, C) Q.2 (AC) Q.3 (B,D) Q.4 (A,B,D)
Q.9 (A,B,C)
Q.10 (A,B,C,D) Q.11 (B,C) Q.12 (B)
Q.13 (D)
Q.15 (C)
Q.16 (B)
Q.19 (B)
Q.20 (A)
Q.21 (AD) Q.22 (B,C) Q.23 (A) p, r (B) p, r (C) q, s (D) q, r
Q.24 (A) q (B) r (C) q (D) r
Q.25 0005 Q.26 0010 Q.27 0050 Q.28 0360 Q.29 0070 Q.30 0270 Q.31 0880 Q.32 4 m/s2 Q.33 0002
EXERCISE-IV
JEE-MAIN
PREVIOUS YEAR'S
Q.1 (1)
Q.11 (1)
Q.2 (4)
Q.12 (1)
JEE-ADVANCED
PREVIOUS YEAR’S
Q.1 (C)
Q.2 (A)
Q.11 (A,C) Q.12 (D)
Q.3 (1)
Q.13 (1)
Q.4 (1)
Q.14 (2)
Q.3 (B)
Q.4 (B)
Q.13 (A, B, D)
Newton’s Laws of Motion and Friction
Q.5 (2)
Q.6 (1)
Q.15 (346) Q.16 (3)
Q.5 (A)
Q.14 (C)
Q.7 (2)
Q.8 (3)
Q.9 (3)
Q.10 (1)
Q.6 (A)
Q.7 10 m/s2 Q.8 (B)
Q.15 (2.00)
Q.9 (A)
Q.10 N = 5
47
JEE - PHYSICS
NEWTON'S LAWS OF MOTION
AND FRICTION
EXERCISE-I
Q.10 (1) m 
Scan for Video Solution
F
62  82  102

 200  10 2 kg.
1

Q.1 (3)
Q.11 (4)
Force
Q.2 (3)
Q.3 (2)
Q.4 (2) Horizontal velocity of apple will remain same but due
to retardation of train, velocity of train and hence
velocity of boy w.r.t. ground decreases, so apple falls
away from the hand of boy in the direction of motion
of the train.
Q.5 (3) Acceleration  
 dv  0.25  10    10  
 m  
 25  20  500 N.
0.01
 dt 
Q.12 (2) F  ma 
m(u  v) 2   8  0 

 4 N.
t
4
Q.13 (3) By drawing the free body diagram of point B
F 100

 20 cm / s2
m
5
C
F
120°
T1
T2
B
120°
120°
Now   t  20  10  200 cm/s
Q.6 (2) u = velocity of bullet
T=10N
dm
= Mass thrown per second by the machine gun
dt
A
Let the tension in the section BC and BF are T1 and
T2 respectively.
= Mass of bullet × Number of bullet fired per second
= 10 g × 10 bullet/sec = 100 g/sec = 0.1 kg/sec
From Lami’s theorem
udm
= 500 × 0.1 = 50 N
 Thrust =
dt
T1
T2
T


sin120º sin 120º sin120º
 T = T1 = T2 = 10 N.
Q.7 (2) Force on particle at 20 cm away F = kx
F = 15 × 0.2 = 3 N
[As k = 15 N/m]
Force 3

 10 m/s 2
 Acceleration =
Mass 0.3
Q.14 (2)
Q.15 (1) For exerted by ball on wall
= rate of change in momentum of ball
 dm 
4
6
Q.8 (3) Thrust F  u 
  5 10  40  110 N
 dt 
Q.9 (3) We know that in the given condition s 
2

58
2
s 2  m1 
m 

  s 2   1   s1
s1  m 2 
 m2 
1
m2

m    m  2mu
.

t
t

Q.16 (1) Given that p  p x ˆi  p y ˆj  2cos tiˆ  2sin tjˆ
 dp
F 
 2sin tiˆ  2cos tjˆ
dt



Now, F.p  0 i.e. angle between Fand p is 90°.
Newton's Laws of Motion and Friction
JEE - PHYSICS
Q.17 (3) Swimming is a result of pushing water in the opposite
direction of the motion.
Q.27 (3)
Q.18 (3)
m2
T '   m1  m 2  
Q.19 (1) Change in momentum = Impulse
 p  F  t  t 
m1
T'
m3
T
T
.
m1  m 2  m3
2m1m 2
2  10  6
Q.28 (4) T  m  m g  10  6  9.8  73.5 N.
1
2
p 125

 0.5 sec .
F 250
Q.20 (1)
Q.29 (3) T sin 30 = 2kg wt
Q.21 (2) F  (F)   F   2F.Fcos     120º.
2
2
T sin 30
30o
T
Q.22 (1) FBD of mass 2 kg FBD of mass 4kg
T
T cos 30
T
4N
30o
T1
8N
2 kg-wt
4 kg
2 kg
20 N
T
T – T' – 20 = 4 ….(i)
T' – 40 = 8
 T = 4 kg wt
T1 = T cos 30º
40 N
= 4 cos 30º  2 3.
…(ii)
By solving (i) and (ii) T= 47.23 N and T = 70.8 N
Q.23 (1)
Q.30 (1) For jumping he presses the spring platform, so the
reading of spring balance increases first and finally it
becomes zero.
Q.24 (2)
Q.31 (3)
Q.25 (2) Tension between m2 and m3 is given by
Q.32 (3) 5N force will not produce any tension in spring without
support of other 5N force. So here the tension in the
spring will be 5N only.
Q.33 (3) As the spring balances are massless therefore the
reading of both balance should be equal.
m1
m2
T
m3
T

2m1m3
g
m1  m 2  m3
2 2 2
 9.8  13 N.
222
Q.26 (2) a  m 2  g  5  9.8  49  5.44 m/s 2
m1  m 2
45
9
Newton's Laws of Motion and Friction
Q.34 (2)
Q.35 (4) As the apparent weight increase therefore we can say
that acceleration of the lift is in upward direction.
R = m (g + a)  4.8 g = 4 (g + a)
ag = 1.96 m/s2
Q.36 (1) T = m (g + a) = 500 ( 10 + 2) = 6000 N
Q.37 (2) Rate of flow will be more when lift will move in upward
direction with some acceleration because the net
downward pull will be more and vice-versa.
Fupward = m (g + a) and Fdownward = m (g – a) Fupward
59
JEE - PHYSICS
Q.38 (1) When car moves towards right with acceleration a
then due to pseudo force the plumb line will tilt in
backward direction making an angle  with vertical.
R
F
a
Q.47 (1)

f
a
60°
F cos 60
F sin 60


g
f = R
F cos 60º = (W + F sin 60º)
From the figure,
tan  = ag
 = tan–1 (a/g).
Substituting  
Q.39 (3)
1
& W  10 3 we get F  20 N
2 3
Q.48 (1) Retarding force F = ma = R =  mg  = g
Now from equation of motion 2 = u2 – 2s
F
F
98
1


 0.1
Q.40 (4)   
R mg 100  9.8 10
 0  u 2  2as  s 
u2
u2
.

2a 2g
Q.41 (1)
Q.42 (3) Fl = sR = 0.4 × mg = 0.4 × 10 = 4N i.e. minimum 4N
force is required to start the motion of a body. But
applied force is only 3N. So the block will not move.
Q.43 (1) For limiting condition  
Q.49 (1) There is no friction between the body B and surface of
the table. If the body B is pulled with force F then
F = (mA + mB) 
Due to this force upper body A will feel the pseudo
force in a backward direction.
mB
5
 0.2 
m A  mC
10  mC
F = mA × 
 2 + 0.2mC = 5  mC = 15 kg
f
Q.44 (3) Sliding friction is greater than rolling friction.
A
B
mB
mB
Q.45 (1)   m  0.2  10  mB  2kg
A
R
F
But due to friction between A and B, body will not
move. The body A will start moving when pseudo force
is more than friction force.
i.e. for slipping, mAa = mAg a = g
Q.46 (3)
fAB
fBG
Q.50 (4)
A
B
F
Ground
Q.51 (2) From the relation F – mg = ma
F = fAB + FBG

= AB mAg + BG (mA + mB)g
= 0.2 × 100 × 10 + 0.3 (300) × 10
= 200 + 900 = 1100 N
60
F  mg 129.4  0.3  10  9.8

 10m/s 2 . .
m
10
Q.52 (3)
Newton's Laws of Motion and Friction
JEE - PHYSICS
R F sin 30°
F
30°
Fk
Q.53 (1)
EXERCISE-II
F cos 30°
Q.1 (4) Experimental fact.
mg
Q.2 (2)
Kinetic friction = kR = 0.2 (mg – F sin 30º)
Q.3(3)
1

 0.2  5 10  40    0.2(50  20)  6N
2

Acceleration of the block

Fcos30º  Kinetic friction
Mass
a
3
6
40 
2

 5.73m/s 2 .
5
Q.54 (1) a = g (sin  –  cos ) = 9.8 (sin 45º – 0.5 cos 45º)

Force exerted by string is always along the string and
of pull type.
When there is a contact between a point and a surface
the normal reaction is perpendicular to the surface and
of push type.
Q.4 (3)
4.9
m / sec2 .
2
6N
2 kg
1 kg
N
3N
Both blocks are constrained to move with same
acceleration.
Q.55 (4) For upper half
6 – N = 2a [Newtons II law for 2 kg block]
2 = u2 + 2al / 2 = 2 (g sin ) l / 2 = gl sin 
N – 3 = 1a [Newtons II law for 1 kg block]
 N = 4 Newton
For lower half
 0  u 2  2g(sin    cos )
l
2
Q.5 (2)
l/2
l/2

– gl sin  = gl (sin  –  cos )
  cos  = 2 sin    = 2 tan 
F – N = Ma [Newtons law for block of mass M]
Q.56 (1) Limiting friction between block and slab = smAg
N – N’ = ma [Newtons law for block of mass m]
= 0.6 × 10 × 9.8 = 58.8 N
N’ = M’a [Newtons law for block of mass M’]
But applied force on block A is 100 N. So the block will
slip over a slab.
Now kinetic friction works between block and slab FK
= kmAg = 0.4 × 10 × 9.8 = 39.2 N
This kinetic friction helps to move the slab
39.2
39.2
2
 Acceleration of slab  m  40  0.98m / s .
B
Newton's Laws of Motion and Friction
 N’ = M’
F
M  m  M'
N = (m + M’)
F
 N > N’
M  m  M'
Q.6 (2) In free fall gravitation force acts.
61
JEE - PHYSICS
T2  3T1
Q.7 (4)
From vertical equilibrium
T2 3 T1
  50
2
2
 T1  25N , T2  25 3N
T cos + T cos – 150 = 0 [Equilibrium of point A]
75
T = cos
When string become straight  becomes 90º
T = 
2 T cos = 150
Q.10 (2) Component of force
in y direction is
NA sin 60° = 500
NA 
1000
3
NA
Q.8 (3)
A
60°
NB
60°
30°
30°
B
Component of force
in x direction is
 NB 
NA cos 60° = NB
500
3
Q.11 (1)
Weighing Machine
always Measure
Normal froce
g
2k
M
W
30º
10 – T2 = 1 a [ Newton’s II law for A ]
T2 + 30 – T1 = 3 a [ Newton’s II law for B ]
T1 – 30 = 3a [ Newton’s II law for C ]
 a=
 T2 =
2k
g
M
30º
N = 20 cos30° = 10 3
g
7
T
6g
7
Q.9 (2) From horizontal equilibrium
T2 T1 3

2
2
T2sin60° T1sin30°
T2cos60°
5Kg.
50 N
62
W
T1cos30°
 T
T
T
Q.12 (2)
m A
B M
C m
T = mg
...(i)
2T cos  = Mg
...(ii)
From equation (i) and (ii)
 2mg cos  = Mg
 always > 0 so M < 2 m
Newton's Laws of Motion and Friction
JEE - PHYSICS
Q.13 (1)At t = 2 sec  a =
So, F = ma =
F – mg = ma2[Newton’s II law for m]
 2 mg – mg = ma2
a2 = g  a2 > a1
10
= 5 m/s2
2
50
 5  0.25 N
1000
At t = 4 sec
a = 0 So F = 0
 At t = 6 sec,
 a = – 5 m/s2  F = – 0.25 N
Q.16 (3) mg –
3
mg = ma [Newton’s II law for man]
4
 a=
g
4
100
100
Q.14 (3)
5kg
M2 g sin – T = M2a [Newton’s II law for M2]
T – M1g sin =
M1a [Newton’s II law for M1]
By adding both equations
M=2.5kg
T
4m/sec2
Q.17 (2)
T+25
70+50
 M 2 sin  – M1 sin  
g
a= 
M1  M 2


120 – 100 = 5a
a=
Q.15 (3) Case 1
....(i)
20
 a = 4m/s2
5
T + 25 – 100 = 2.5 × 4
T = 85 N
T1
2
2
2
Q.18 (4)   0  2as = 1  2
T1
x=
T1
T1
mg
F
x
m
–m 2
  02  2as
2F
2F1  – m 
2F1
 0=9+


O =3 +
m  2F 
m
a1
a1
....(ii)
2
2
 F1 = 9F
2 mg
Q.19 (3)
T1 – mg = ma1 [Newton’s II law for m]
2 mg – T1 = 2 ma1 [Newton’s II law for 2m]
 a1 =
g
3
Case 2
F
By dividing both equations
a2
F
F = 2 mg
Newton's Laws of Motion and Friction
Mg sin – T = Ma [Newton’s II law for block 1]
T = Ma [Newton’s II law for block 2]
mg
2 T = Mg sin T =
Mg sin 
2
63
JEE - PHYSICS
T
Q.20 (3)
m1
T1 – 12 g – T2 = 12 a
T
a
m2
 T1 = 12 × 2.2 + 12 × 9.8 + 96
T1 = 240 N
a
Q.24 (1) -vA -vA-vA + vB = 0
m2g
m1g
From constrained
– 5 – 5 – 5 + vB= 0
T – m1g = m1a
....(i)
m2g – T = m2a
....(ii)
On solving equation (i) and (ii)
vB = 15 m/s 
Q.25 (1)From constrained
 m  m1 
a  2
g
 m1  m 2 
Q.21 (1)
+2 – vB – vB + 1 = 0
vB  3 / 2 m / s 
T1
A
T2
a
mAg
B
v
Q.26 (4)
a
A
mBg
u 
B

u
Q
P
T2
T1  T2  m A g  m A a
M
T2  m B g  m B a
From constrained
The resultant vel. of the Block M is v in vertical
direction.
So component of 'v' is direction of u is
T1  (mA  m B )g  (m A  m B )
T2  m B (a  g)
T1 m A  m B

T2
mB
T = 250 (max)
Q.22 (4) Tmax = mmaxg = 25 × 10 = 250
250 – 200 = 20 amax
amax = 2.5 m/s2
Q.23 (3)
u
cos 
amax
20g
Q.27 (4)
T1
12g
a = 2.2 m/s
2
V = (velcoity of B w.r.t ground)
T2
V–4
= 2V = 8 m/s (velcoity of B w.r.t ground)
2
T2
8g
a = 2.2 m/s2
T2 – 8g = 8a [ Newton’s II law for 8 kg block]
 T2 = 8 × 2.2 + 8 × 9.8
= 96 N
64
v cos  = u  v =
V' = 6 m/s (velcoity of B w.r.t lift )
Q.28 (2) 18kg at rest => 180 = 2F
F = 90N
Newton's Laws of Motion and Friction
JEE - PHYSICS
Q.33 (3)
a
Q.29 (3)
T1
F
F
F
T2
6 kg
T1
T2
3 kg
1 kg
a
18kg
C
mg = 10
(a) T = mg + ma (b) T = mg – ma
T = mg
a
mg = 30
30 – T2 = 3 a [Newton’s II law for 3 kg block]
T2 – T1 = 6 a [Newton’s II law for 6 kg block]
T1 – 10 = 1 a [Newton’s II law for 1 kg block]
By adding three equations
30 – 10 = 10 a  a = 2 m/s2.
Q.30 (1)
2m1m 2 g
2  5  1 10
50
=
Q.34 (2) T  (m  m )  T 
6
3
1
2
T – mg = 0 [ Equilibrium of block]
T – 10 = 0
T = 10
Reading of spring balance is same as tension is spring
balance.
Q.31 (4)
a1
F
a2
kx
kx
M1
kx
2T 
The spring balance reads
2T = 33.33kgwt < 60kgwt
Q.35 (2)
2ms
3 kg
10 – kx = 2 × 2
kx = 6 N
F – k x = m1 a1 [Newton’s II law for M1]
kx = m2a2 [Newton’s II law for M2]
By adding both equations.
F – m1 a1
F = m1a1 + m2a2  a2 =
m2
–2
2 kg
10N
2 m/s2
M2
kx
100
 33.3kg
3
kx
2kg
10N
 Acceleration of 3 kg =
6
= 2 m/s2
3
Q.36 (4) Weight of man in stationary lift is mg.
N
Q.32 (3)
a
2 kg
T
T
T
Reading of spring balance is same as tension in the
balance.
 T = 10 g = 98 N
T = 2 a [Newton’s II law for 2 kg block]
 a = 49 m/s2
Newton's Laws of Motion and Friction
mg – N = ma [Newton’s II law for man]
 N =m (g – a)
Weight of man in moving lift is equal to N.
65
JEE - PHYSICS
L
 1
Q.41 (2) L 1   g   g
n
 n
mg
3
 a= g
m (g – a) 2
3

Q.37 (1) F < fsmax
OOOOOOOO
OOOOOOOOO
F
f
friction=F
For F > fmax
friction constant
Q.38 (1) Monkey is moving up due to friction force
L/m
1
n 1
Q.42 (1)N = mg cos 
fs   N
mg sin   m g cos 

f
N
mg cos 
mg sin 
fr – mg = ma

fr = m(a+g)
towards up.
 1
Q.39 (3) Floor will provide the normal force and friction force
the net reaction is provide by the floor is R.
N
Q.40 (2)
V=
A
fr
R
A
so angle remain same.
 Angle = 30°
V
floor
a
fmax > mg sin
sin
at this condition block remains rest when
mg sin > fmax
sliping slant
FC
N
f
N
F
Fsin
m
Q.44 (3)
mg
fr
F sin  + N = mg
or N = mg – F sin  ...(1)
m
mg
For  < angle of repose
Fc = mg
For  > angle of repose
as   f = mg cos  
N = mg cos  
66
Q.43 (1)Friction not depend on surface Area
fr = N
...(2)
F cos  – fr = ma
...(3)
on solving (1), (2) & (3)
a
F cos   (mg  Fsin )
m
a=
F
(cos + sin) – g
m
Newton's Laws of Motion and Friction
JEE - PHYSICS

(B)

(C)

(D)
V 2  2  g sin  l
Q.45 (1)
v2
n2
 2  (g sin   g cos )
1 

sin  1  2    cos 
 n 
Acceleration of m2 is 
T = m2 g > m1g
acceleation of m1 is 
Masses is different
Not possible
T – m1g = m1a
m2g – T = m2a
(m 2  m1 )g
on solving a  (m  m ) Possible
1
2
1 

  tan  1  2 
 n 
Q.5 (ABD)
EXERCISE-III
Q.1 (A, B, C) F = t
dv

=
t ....(i)
a=
dt
m
By string constraint
straight line curve 1 dv =
 t2
v=
curve 2
m 2
divide (ii) by
v=
aA = 2aB ................................(1)

tdt
m
equation for block A.
...(ii)
10 × 10 ×
t
a am
a 2m
a = 
=
 Paacebole curve 2.
2
2 
2
F
F = 2 T cos  T =
2 cos 
Q.2 (AC)
equation for block B.
2T –
a
A
fixed
T
T
T
B
mg
30°
T + mg sin  = ma
mg – T = ma
....(1)
....(2)
on solving (1) & (2)
a=
T=
Q.4 (A,B,D)
3g
4
400
= 40 aB
2
.............(3)
Solving equation (1) , (2) & (3) we get
  cos   T 
on incrcasing , cos decreases and hence T increases.
Q.3 (BD)
1
– T = 10 aA ......(2)
2
aA =
5
m/s2
2
aB =
5
m/s2
2 2
T
150
N
2
=
3g
4
Q.6 (AC)
T
T
m1g
m2g
(A) T = m1g < m2g
Newton's Laws of Motion and Friction
T = m1g
when thred is burnt, tension in spring remains same =
m1g.
m m
m1g – m2g = m2a  m1  m2  g = a = upwards
 1
2
67
JEE - PHYSICS
(C) f c  f 2  N 2  (100) 2  (200) 2  100 5 N
Q.11 (B,C)
T
a
T
m
7° 100
in3 f
s
mg
37°
for m1
mg
T = mg
T = 100 mg sin 37° + 0.3 × 100 g cos 37°
[Put g = 9.8]
T = 588 + 235.2
mg = 823.2  m = 82.33 = 83 kg
a=o
T
a
Q.7 (BC)
(b)
Apply NLM on the system
200 = 20 a + 12 × 10
g
m
80
= a = 4 m/s2
20
spring Force = 10 × 12 = 120 N
Q.8 (CD) Pseudo force depends on acceleraton of frame and
mass of object
Q.9 (A,B,C)
T
0
10
°
37
n
i
s
m
mg
T + f = ma
T + 235.2 = 588
T = 588 – 235.2 = 352.8
m = 35.28 kg
Q.12 (B) FBD of Block in ground frame :
applying N.L.
150 + 450 – 10 M = 5M
 15 M = 600  M =
M
f
T
600
15
150 N
N
450 N
T – N = ma
As T  man
Can have tendency
to move
5 m/s2
Mg = 10 M
  0.6
Q.10 (A,B,C,D)
10 kg
20 m/s2
(A) Acceleration of box = 20 m/s2
(when consider as system)
Force on Box
F = 200 N
N = 200 N
fmax = N = 0.6 × 200 = 120 N
(B) frequired = mg = 10×10 = 100 N
68
 M = 40 Kg Ans.
Normal on block is the reading of weighing machine
i.e. 150 N.
Q.13 (D) If lift is stopped & equilibrium is reached then
T = 450 N
N
450 + N = 400
Mg = 400 M
Newton's Laws of Motion and Friction
JEE - PHYSICS
 N = – 50
So block will lose the contact with weighing machine
thus reading of weighing machine will be zero.
T
Q.19 (B) S2 is inertial frame
F = ma
So F  10iˆ  20ˆj
Q.20 (A) With respect to S1 frame
Net force = zero.
T = 40 g
Q.21 (AD)
40 g
So reading of spring balance will be 40 Kg.
(A) When F = 0
No frictin b/w m & M0 so system move.
(B) When F is applied then friction develope a range
for which M and m are stationary w.r.t M0, such that
Mg = 400 N
950  400
40
450
45
a=
=
m/s2 Ans.
4
40
a=
t

f
Putting v = 2 we have t = 2 sec.
2
Q.22 (B,C)
4
=
3
4
8
=
m
3
3
Q.16 (B) From above
2t =
t3
 t2 = 12  t = 2 3 sec.
6
Q.17 (C) a = t = 4  after 4 seconds VB =2 m/s
Vp =
42
= 8 m/s  Vrel = 8 – 2 = 6 m/s.
2
Q.18(A) S1 is accelerating frame so psuedo force act opposite
to frame acceleration
FPseudo = mass of analyzing body × acceleration of frame
 2(–5iˆ – 10ˆj) = –10iˆ  20 ˆj
Newton's Laws of Motion and Friction
mg
F
M
T
Mo
T
m
mg
f
xB = 2 × 2 = 4 m
Hence relative displacement = 4 –
mg
a
Ma
 t3 
 
 6  0
m
(C) Limiting friction between M0 & m is  ma
 Dependent on a
(D) When Pseudo acts on M is equal to T then f = 0
2
dv  t dt  v = t
2
0
0
dx t 2
  xp =
Now
dt 2
or
m
f

T
T
10 t
=t
Q.15 (C)ap =
10
v
m
Rough
N = 400 N
a
dv

=t
dt
T
Mo
F
T = 450 N
Q.14 (A) 40 Kg
T
Smooth M
Use Pseudo concept
T = Ma
T = f + mg
 T =  ma  ma
On using (1) & (2)
Ma =  ma + mg
...(1)
...(2)
Q.23 (A) p, r (B) p, r (C) q, s (D) q, r
(A) Let the horizontal component of velocity be ux.
Then between the two instants (time interval T) the
projectile is at same height, the net displacement (uxT)
is horizontal
 average velocity =
uxT
= ux (A) p, r
T
(B) Let î and ĵ be unit vectors in direction of east and
north respectively.
69
JEE - PHYSICS



 V DC  20 ˆj , V BC  20iˆ and V BA  20 ˆj




 VAD  VDC  VCB  VBA = 20 ˆj  20 ˆi  20 ˆj
=  20 ˆi

F2  F1
a= m m
1
2


 VAD  20iˆ Hence VAD  VBC (B) p, r
(C) Net force exerted by earth on block of mass 8 kg is
shown in FBD and normal reaction exerted by 8 kg
block on earth is 120 N downwards.
The FBD of m2 is
 F2 – N2 = m2a
m1 m 2
Solving N = m  m
1
2
(D) Replace F1 by –F1 in result of C
m1 m 2
N= m  m
1
2

Hence both forces in the statement are different in
magnitude and opposite in direction. (C) q,s
(D) For magnitude of displacement to be less than
distance, the particle should turn back. Since the
magnitude of final velocity (v) is less than magnitude
of initial velocity (u), the nature of motion is as shown.
 Average velocity is in direction of initial velocity
uv
is
and magnitude of average velocity =
2
less than u because v < u. (C) q, r
Q.24 (A) q (B) r (C) q (D) r
Let a be acceleration of two block system towards right
F2  F1
a= m m
1
2
The F.B.D. of m2 is
 F1
F 
 2 

 m1 m 2 
 F2
F 
 1 

 m 2 m1 
Q.25 [0005 ] T – mg = ma
T
.......... (i)
2T
m
a
m
mg
a'
4mg
4mg – 2T = 4ma'  2mg – T = 2ma'..... (ii)
constraint relation : Ta – 2Ta'  a' = a/2
 from equation (ii) we get
2mg – T = 2m (a/2)
........ (iii)
solving (i) and (ii), we get
a=
g
= 5 m/s2
2
Q.26 [0010] T1 sin 30° + T2 sin 60° = 20
30°
T1
60°
T2
2g
T1 cos 30° = T2 cos 60°
 F2 – T = m2 a
m1 m 2  F2
F1 
Solving T = m  m  m  m 
1
2 
2
1 
(B) Replace F1 by – F1 is result of A
m1 m 2
T= m m
1
2
 F2
F 
 1 

m
m
 2
1 
(C) Let a be acceleration of two block system towards
left
70
T2 = T1
3

4T1
= 20
2
T1 = 10 N
Q.27 [0050 ]
100 = k × (40 – 0)
200 = k × (60 – 0)
2=
60   0
40   0
80 – 20 = 60 – 0
20 = 0


k=5
x = k × (30 – 20) = 50 N
20k = 100
Newton's Laws of Motion and Friction
JEE - PHYSICS
Q.28 [0360]
90  50
= 8, after 2 sec
5
a=
S = 16m, V = 16 m/s
3 kg part hits the ground t = 4 sec.
remaining 2 kg has a = 35
Srel = 360
N
Q.34 [0200] N
F cos 37° = N
F sin 37° + N = 100 g
–––––––––––––––––
T
F=
mAg
mBg
7T = mAg
T = mBg
 mA = 7mB = 70
Q.30 [0270]
37°
 a=0
100g
Q.29 [0070]
4T 2T T
F
µ mg
cos    sin  = 200 N
EXERCISE-IV
JEE-MAIN
PREVIOUS YEAR’S
F = 2 × 75 cos 37° + 150
= 150 ×
4
+ 150 = 270
5
Q.1 (1)
Q.31 [0880 ] m = 16, M = 80, x = 6, L = 12

 mx
 M g

 L
T= 
16  6

 12

 80  10 = 880 N

T
a
Q.32 [4]
50g
50g – T = 50a
a = 10 –
300
= 4 m/s2
50
Q.33 [0002]
amax = µg = 0.2× 10 = 2m/sec2
v = u + at
0 = 4 – 2t
t = 2sec
Newton's Laws of Motion and Friction
50 – T = 5 × a
T – 0.15 (m + 10) g = (10 + m)a
a = 0 for rest
50 = 0.15 (m + 10) 10
5=
3
(m + 10)
20
100
= (m + 10)
3
m = 23.3 kg
Q.2 (4) T cos 45° = mg
T sin 45° = mg

F = mg 100 N.
Q.3 (1) For equilibrium of the block net force be zero. Hence
we can write.
mg sin  + 3 = P + friction
mg sin  + 3 = P + mg cos 
After solving , we get, P = 32 N.
Q.4 (1) mg sin  + 2 = µ mg cos 
10 – mg sin  = µ mg cos 
On adding (1) + (2)
12 = 2µmg
....(1)
....(2)
12
3
; µmg =
3
2
71
JEE - PHYSICS
On (1) – (2) :
1
8 = 2mg × ;
2
10
3
mg = 8 ; µ =
2
20
N2
30º
10 3
f k2
50
dv
dt
Q.5 (2) –(g + v2) =
N1 = 60



g
dv 


– gdt =   g
2 

v




Intergrating 0  t & V0  0 :-
–gt = –
t=



V 
g
tan 1  0 

 g
 


10 3  0.2  60
5
a2 
10 3  0.2  40
5
Q.8 (3)
T 45
o
F
100N
100N
10kg
  
1
tan 1 
V0 
g
 g 
T
 100
2
T
F
2
1kg
3kg
F = 100 N
F
aAmax = g = 2 m/s2
1kg
3kg
(0.2)4g
=8N
F–8=4×2
F = 16 N
30º
f k1
50
Q.9 (3)
|a| = g + kv2
F
vdv
 g  kv 2
dh
–


0
u
vdv
g  kv
2


H max
0
dh
On solving
N1
72
a1 
a1–a2 = 0.8
Q.6 (1)
Q.7 (2)
N2= 40
10 3
 ku 2
1
Hmax = 2K ln 1  g




20
Newton's Laws of Motion and Friction
JEE - PHYSICS
N1
Q.14 (2)
10 3
30º
f k1
50
Q.10 (1)
20
10
20
N2
30º
f k2
10 3
50
50 – T = 5 × a
T – 0.15 (m + 10) g = (10 + m)a
a = 0 for rest
50 = 0.15 (m + 10) 10
N1 = 60
a1 
N2= 40
10 3  0.2  60
5
10 3  0.2  40
5
a1–a2 = 0.8
3
(m + 10)
5=
20
a2 
100
= (m + 10)
3
m = 23.3 kg
consider this to be 27.3.
Q.15 (346)
For upward motion
s
Q.11 (1) For equilibrium of the block net force be zero. Hence
we can write.
mg sin  + 3 = P + friction
mg sin  + 3 = P + mg cos 
After solving , we get, P = 32 N.
=30°

1
S
3 
mv02   mg  µmg
S
2
2
2 

Q.12 (1) mg sin  + 2 = µ mg cos  ....(1)
10 – mg sin  = µ mg cos  ....(2)
On adding (1) + (2)
3
;
2
On (1) – (2) :
12 = 2µmg
µmg =
1 mv02 
S
3 
  mg  µmg
.S 

2 4
2
2 

12
3
For downward
4=
1
8 = 2mg × ;
2
3
mg = 8 ; µ =
2
(1  µ 3)
 4 – 4µ 3  1  µ 3
3 346.41

 3 = 5µ 3  µ =
5
1000
I = 346
1kg
Q.13 (1)
(1  µ 3)
3kg
F
Q.16 (3)
y 
aAmax = g = 2 m/s2
(0.2)4g
=8N
F–8=4×2
F = 16 N
Newton's Laws of Motion and Friction

h
 = tan
1kg
3kg
R = 1m
F
3
= tan
4
  = 37º

 h = R – R cos = 1 – 1 ×
4
= 0.2 m
5
73
JEE - PHYSICS
JEE-ADVANCED
PREVIOUS YEAR’S
Q.1 (C) 2mg cos  =
1
cos  =
2
Q.5 (A)
2 mg
= cos 45°   = 45°
Q.2 (A) After string is cut, FBD of m
N = mg + F sin 60 = 3 × 10 +
F cos 60 = N
mg
= g
m
FBD of 2m (when string is cut tension in the spring
takes finite time to become zero. How ever tension in
the string immediately become zero.)
a=
3mg
a=
Q.3 (B)
2m
3mg  2mg
g
=

2m
2
2mg
a
F cos 
= 2m sin 
=
Q.4 (B)
F
2m

F
F
=5+
2
4
=
........ (ii)
F 3
)
× (10 3 +
2
2 3

8
Q.6 (A) aA = g [sin 45 – A cos 45] =
2
F
=5
4
,
7
aB = g [sin 45 – B cos 45] =
T cos 
m
=
F
2
...... (i)
 F = 20 N
F = 2T sin 
a
1

F 3
2
2
aAB = aA – aB = g (B – A) cos 45 =
1
2
,
sAB = 2
x
2
a x
1
2
Now sAB =
2
ma cos  = mg cos (90 – )

2
=
Again sA =
1
2
1
2
aAB t2
×
1
2
aA t2 =
t2
1
2
 t = 2 sec.
(
8
2
)4
 sA = 8 2 m

a
a
dy
 tan  
g = dx
g

a
d
(kx2) = g
dx
a
 x = 2gk = D
74
Q.7 [10]
Solving from the frame of disc
Newton's Laws of Motion and Friction
JEE - PHYSICS
F1 =
mg mg
mg mg


;F2 =
2
2
2
2
F1 = 3F2
1 +  = 3 – 3
4 = 2
=
Let accleration of the block relative to the disc is a then
25 m cos  – N = m a
..........(i)
Now, there will be two normal as there are two contacts
1
2
N = 10
N=5
Q.11 (AC)
(i) Horizontal and (ii) vertical
NH = 25 m×sin = 25 × m ×
3
= 15 m
5
NV = mg = 10 m
f = NH + NV =
2
2
(15 m) +
(10 m) = 10 m
5
5
from (i) we get
a=
m (25 cos  – 10)
= 10 m/sec2
m
Q.8 (B) Statement-1 is also practical experience based; so it
is true.
Statement-2 is also true but is not the correct
explanation of statement-1. Correct explanation is
''there is increase in normal reaction when the object
is pushed and there is decrease in normal reaction
when object is pulled".
f = 0, If sin = cos
f towards Q, sin > cos
f towards P, sin < cos
  = 45°
  > 45°
  < 45°
Q.12 (D)
Block will not slip if
(m1 + m2) g sin m2 g cos
 3 
 (2) cos
 10 
3 sin  
Q.9 (A)
tan 
P1 = mgsin – mgcos
P2 = mgsin + mgcos
Initially block has tendency to slide down and as tan
> , maximum friction mgcos will act in positive
direction. When magnitude P is increased from P1 to
P 2, friction reverse its direction from positive to
negative and becomes maximum i.e.mgcos in
opposite direction.
(P)  = 5°
f = (m1 + m2)g sin
(Q)  = 10°
f = (m1 + m2)g sin
(R)  = 15°
f = m2g cos
(S)  = 20°
f = m2g cos
Q.13(A, B, D)
Q.10 [5]
Newton's Laws of Motion and Friction
1
5

dk
 t as
dt
   11.5
friction is static
friction is static
friction is kinetic
friction is kinetic
1
k  mv 2
2
dk
dv
 mv  t
dt
dt
75
JEE - PHYSICS
v
t
0
0
1
 vdv  14m  (8Mg  3kx)dx
 m  vdv    tdt
for max elongation
x
mv 2 t 2

2
2

t
m
v
.....(i)
=
x0 =
since F = ma

F  m
 m  constant
m
at x =
x0
2
 vdv 
Q.14 (C)
kx
2m
2m
a1
0
2T
2T
T
m
a1
T
2m
T=
2(2m)(m)
(g – a1 )
3m
2T – kx = 2ma1
3kx 20
2
16Mg
3k
v
1
14m
x0 / 2

(8Mg  3kx)dx
0
v2
1  8Mgx 0 3kx 02 




2 14m  2
2 4 
1  8Mg 16Mg 3x 16M 2 g 2 

 


v =
7m  2
3x
8 3x  3x 
2
=
1  64M 2 g 2 2M 2g 2 



7m  3x
3x 
62Mg 2
21k
For acc. 2a1 = a2 + a3 therefore
a2 – a 1 = a 1 – a 3
v2 =
8Mg  3kx 0 / 4
14m
4m
=
(g – a1)
3
a1 =
8m
(g – a1 ) – kx = 2ma1
3
=
8g
3kx 0

14 14m  4
8Mg 8ma1

 kx  2ma1
3
3
=
8g
3x
16Mg


14 14m  4
3x
8Mg
14ma1
 kx 
3
3
=
8g 4g

14 14
8Mg  3kx
= a1
14m
=
a1 =
8mg  3kx
14m
vdv  8Mg 3kx 



dx  14m 14m 
76

3kx 02 
 8Mgx 0 

2 

1
14m
8Mgx0 =

dv

 constant
dt
m
a
1 0
(8mg  3kx)dx
14m 0
0=
4g 2g

14 7
OR
8mg 8m
–
a – kx = 2ma1
3
3 1
 8mg 
14m
a1 = –k  x 
3k 
3

Newton's Laws of Motion and Friction
JEE - PHYSICS
3k  8mg 
x
.....(i)
3k 
14m 
that means, block 2m (connected with the spring) will
a1 = –
8mg
perform SHM about x1 =
therefore.
3k
maximum elongation in the spring x0 = 2x1 =
on comparing equation (1) with
a = –2(x– x0)
=
16mg
3k
 x0 
at   , block will be passing through its mean posi 2
tion therefore at mean position
At.
3k
14m
A
x0
x=
2
4
 acc = –
=–
8mg
3k
A 2

2
4mg 3h
2g
.

3k 14m 7
Q.15(2.00)Let TS = tension in steel wire
TC = Tension in copper wire in x direction
30º
TS
TS cos60º
3k
14m
v0 = A =
60º
TC
60º
30º
x
TC cos60º
100
TC cos30º = TS cos60º
TC ×
1
3
= TS ×
2
2
3 TC = TS
in y direction
TC sin30º + TS sin60º = 100
TC TS 3

= 100
2
2
Solving equation (i) & (ii)
TC = 50 N
......(i)
.....(ii)
TS = 50 3 N
We know
L =
FL
AY
LC
TC LC AS YS
= L = A Y  T L
S
C C
S S
On solving above equation
L C
LS = 2
Ans. 2.00
Newton's Laws of Motion and Friction
77
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