NEWTON’S LAWS OF MOTION AND FRICTION
Single Type
1.
Find the angle   90º, such that the block of mass m does
not move on the rough surface for any value of the applied
for F.
F m

(
)R
ou
ghS
urface
(A) tan1()
(B)
1
tan 1  
2
(C) cot1()
(D)
1
cot1  
2
Ans: (C)
…..(i)
…..(ii)
F cos   N
N = F sin + mg
By (i) and (ii)
F cos  F sin   mg
Fcos   sin   mg
if F is . Hence
cos    sin   0
 cot  = 
  = cot1()
cos    sin   0
2.
A body of mass ‘m’ is released from rest on an inclined
plane in an elevator (moving up with acceleration a0). The
time taken by the body to reach the bottom of the inclined
plane is
1
a0
m
sm
(A)
L
oo
th

L
a 0  g sin 
(C)
(B)
3L
a 0  g sin 
(D)
2
t
2L
a 0  g sin 
L
a 0  g sin 
Ans: (B)
From the free body diagram, we get,
N  mg cos   ma 0 cos 
mg sin   ma 0 sin   ma
a  a 0  g sin 
L
1
 a 0  g  sin  t2
2
t
3.
2L
 a 0  g  sin 
What is the minimum value of the mass M so that the block
is lifted off the table at the instant shown in the diagram?
Assume that the blocks are initially at rest.
Frictionless
Table
m 60
o
M
(A)
m
sin 60
(B)
(C)
m sin 60
(D) none of these
m
tan 60
2
Ans: (D)
The accelerations of the blocks along the string are equal;
T
T
60°
m
M
a1
mg
a
Mg
a  a1 cos 60
or a  2a
For the block to lift up
(1)
T sin60  mg
(2)
1
Also,
(3)
(4)
T cos 60  ma1
Mg  T  Ma
 M
4.
4M
2 3 1
Assuming all the surfaces to be frictionless find the
magnitude of net acceleration of smaller block m with
respect to ground
M m
(A)
2 5mg
(5m  M)
(B)
(C)
7 5mg
(5m  M)
(D) none of these.
2mg
(5m  M)
Ans: (A)
F.B.D. of m
3
T
m
a
N
mg
2a
F.B.D. of M
N’
T
M
N
T
mg
For m
mg – T = m.2a …(i)
N = ma
…(ii)
For M
2T – N = ma
…(iii)
On solving
a
2mg
(M  5m)
 Net acceleration of m,
a m  4a 2  a 2  5a

5.
2 5mg
.
(5m  M)
The minimum value of
slipping is

between the two blocks for no
m

2m
F
smooth
4
(A)
(C)
F
mg
2F
3mg
(B)
(D)
F
3mg
4F
3mg
Ans: (C)
For no slipping am = a2m
F  mg  ma
F
…(1)
2m
a 
a
m
mg
 mg
mg = 2 ma
...(2)
g
2
Putting in (i)
F   mg 
mg
2
3
 F  mg
2
or   2 F
3mg
6.
A particle is placed at rest inside a hollow hemisphere of
radius R. The coefficient of friction between the particle
and the hemisphere is   1 . The maximum height upto
3
which the particle can remain stationary is
(A)
R
2
(B)

3
1 
 R
2


(C)
3
R
2
(D)
3R
8
Ans: (B)
5
N  mg cos 
f max  mg sin 
N  mg sin 
1
[mg cos ]  mg sin 
3
1
 tan  
or   30
3
Hence
7.
h  R  R cos 30  R 

3
3
R  R 1 

2
2 

Assuming all the surfaces to be frictionless find the
magnitude of net acceleration of smaller block m with
respect to ground
M m
(A)
2 5mg
(5m  M)
(B)
(C)
7 5mg
(5m  M)
(D) none of these.
2mg
(5m  M)
Ans: (A)
F.B.D. of m
T
m
a
N
mg
2a
F.B.D. of M
6
N’
T
M
N
T
mg
For m
mg – T = m.2a …(i)
N = ma
…(ii)
For M
2T – N = ma
…(iii)
on solving
a
2mg
( M  5m)
 Net acceleration of m,
am  4a2  a2
8.
 5a 
2 5mg
(5m  M )
.
A lift is moving in upward direction with an acceleration 2
m/s2. A body is projected in horizontal direction with initial
velocity 12 m/s. The coefficient of friction between body
and floor of the lift is  = 0.5. Find the displacement of
body, when it comes to rest w.r.t. lift
a
v
(A) 12 m
(C) 4 10m
Ans: (C)
(B) 20 m
(D) 4m
7
N
v
f=  N
m(g+a)
In the frame of lift the normal reaction N  m  g  a 
 the friction force f  N  m  g  a 
So if the body moves a horizontal distance x then
1
mv 2  m  g  a   x
2
12 
v2
=
x

2  g  a  2  0.5  10  2 
2
12m
and if the time taking by the body for this movement is t
then by equation of motion
v  u  at
v0
u  12 m/s
0  12  6  t
a    g  a   0.5  10  2   6 m / s 2
t = 2 sec.
But lift also moves upward in this duration. So the vertical
displacement of body is
y
1 2 1
2
at   2   2   4
2
2
Total displacement w.r.t. ground
 x 2  y2 
9.
12    4 
2
2
 160  4 10 m .
A particle is hanging from a fixed point O by means of a
string of length ‘a’. There is a small smooth nail O in the
same horizontal line with O at a distance b(  a) from O.
8
The minimum velocity with which particle should be
projected from its lowest position in order that it make a
complete revolution round the nail without string become
slack
O'
O
b
a
(A) 3ga
(B) 5ga
(C) (5a  3b)g
(D) (5b  3a)g
Ans: (C)
Let the velocities at point A & B are vA & vB respectively,
using energy conservation between points A and B
B
(a-b)
O'
O
b
O
O'
b
a
a
A
A
1
1
mv 2A  mv B2  mg a  a  b
2
2
v2A  vB2  2g  2a  b 
To complete the circle about
O v B  g  a  b 
v2A  g  a  b   2g  2a  b   ag  bg  4ga  2gb  5ga  3gb
v2A  g  5a  3b 
vA  g
 5a  3b  .
9
10. In the figure, the wedge is pushed with an acceleration of
10 3 m / s . It is seen that the block starts climbing upon the
smooth inclined face of wedge. What will be the time taken
by the block to reach the top?
2
1m
m
a = 10 3 m/s2
30°
(A)
2
s
5
(B)
1
s
5
(C)
5s
(D)
5
s
2
Ans: (B)
3
 15 m / s 2
2
10 3 cos 30  10 3
10 3 m/s2
30°
g sin30° = 5 m/s2
 a  15  5  10 m / s 2
S 
1 2
at
2
1
1  (10)t 2
2
or
t
1
S
5
Multiple Correct Type
11. Co-efficient of friction for all surfaces is  . The strings and
pulleys are ideal. Blocks are moving at a constant speed.
Choose the correct options
A
F
T1
2m
3m
B
T2
10
(A) F  9mg
(C) T  6mg
Ans: (A, B)
(B)
(D)
2
T1  2mg
T1  4mg
N1
N1
A
T1
2mg
N
1

N
1
T
T
2=
1
B
F
N
2

N
2
3
m
g
As string is ideal T1 = T2
T1 = N1 = 2 mg
F = T1 + N1 + N2 = 2 mg + 2mg + 5mg = 9 mg.
12. In the figure, the pulley P moves to the right with a
constant speed u. The downward speed of A is A, and the
speed of B to the right is B
B
P
u
A
(A)   
(B)   u  
(C)   u  
(D) the two blocks have accelerations of the same
magnitude.
Ans: (B, D)
A
B
B
A
B
A
11
At any instant of time, let the length of the string BP = l1
and the length PA = l2. In a further time t, let B move to the
right by x and A move down by y, while P moves to the
right by ut. As the length of the string must remain
constant,
l1 + l2 = (l1 – x + ut) + (l2 + y)
or x = ut + y
or x  u  y
x = speed of B to the right = vB,
y = downward speed of A = vA
.
.
.
.

vB  u  vA
.
.
Also, v  v
or, aB = aA.
B
A
13. A block of mass m1 is connected with another block of
mass m2 by a light unextended spring kept on a smooth
horizontal plane. m2 is connected with a hanging mass m3
by an inextensible light string. At the time of release of
block m3:
m1
m2
m3
m 2 m3
m 2  m3
(A) tension in the string is
g
(B) acceleration of m1 is zero
(C) acceleration of m2 is m mm g m
3
1
2
3
12
(D) acceleration of m2 is
m3
g
m3  m 2
Ans: (A, B, D)
Initially spring is unstretched
a m1  0
a m2 
m 3g
m 2  m3
m3g  m 2 a  m3a
T
m 2 m3
g.
m 2  m3
14. Two masses A and B lie on a frictionless table. They are
attached to either end of a light rope which passes around a
horizontal movable pulley of negligible mass.
mA = 1 kg, mB = 2 kg, mc = 4 kg. Pulley P2 is vertical.
B
P1
P2
A
C
(A) aA = 8 m/s2
(C) ac = 6 m/s2
Ans: (A, B, C)
(B) aB = 4 m/s2
(D) aA > aB > aC
2
a
c
2
T
B
T
A
T
a
c
4
0
FBD of A and B
w.r.t. pulley P1
Equations of motion are
13
T – 2ac = 2a1
ac – T = a 1
and 40 – 2T = 4a
Solve to get the answers.
…(i)
…(ii)
…(iii)
15. The acceleration of blocks of masses 5 kg and 10 kg are
F
A
a1
B
a2
5kg
10kg
(A) zero if F  100N
(B) a  5m/ s and a  0 if F  300N
(C) a  15m/ s a  2.5m/ s if F  500N
(D) Acceleration of the masses is independent of F
Ans: (A, B, C)
2
1
2
2
1
2
2
F
A
T'
T'
aT
1
5kg
2T '  F
;
T' 
F
2
;
B
T
T
a2
T
10kg
2T  T '  F / 2 ; T  F / 4
F
 5g  F  200 N (5kg is lifted off)
4
F
 10g  F  400 N(10kg is lifted off)
4
14
When F = 100 N
a1  a2  0
When F = 300 N
F
a  0 ;  50  5 a  a
2
1
4
1
 5m / sec 2
When F = 500 N
F
 50
4
a1 
 15m / sec 2
5
;
F
 100
4
a2 
 2.5m / sec 2
10
16. A man in a lift ascending with an upward acceleration a
throws a ball vertically upwards with a velocity v and
catches it after t second. Afterwards, when the lift is
descending with the same acceleration a acting downwards,
the man again throws the ball vertically upwards with the
same velocity and catches it after t seconds. Which of the
following statements are correct?
(A) The velocity of the ball is g  t  t  / t t
(B) The velocity of the ball is gt t /  t  t 
1
2
1
2
1 2
1 2
1
2
(C) The acceleration of the ball is g  t
 t1 
 t 2  t1 
(D) The acceleration of the ball is
2
t
t 
g 1  2 
 t 2 t1 
Ans: (B, C)
t1 
2v
2v
or , g  a 
ga
t1
t2 
2v
2v
or , g  a 
ga
t2
Adding,
2g 
2v 2v

t1
t2
15
or,
t t 
v 1 2   g
 t1t 2 
 gt t 
v 12 
 t1  t 2 
2 gt t
2gt 2
ga   1 2 
t1 t1  t 2 t1  t 2
or
a
g  t 2  t1 
 t 2  t1 
17. In the figure, a man of true mass M is standing on a
weighing machine placed in a cabin. The cabin is joined by
a string with a body of mass m. Assuming no friction, and
negligible mass of cabin and weighing machine, the
measured mass of man is (normal force between the man
and the machine is proportional to the mass)
m
(A) measured mass of man is
Mm
(M  m)
(B) acceleration of man is
mg
(M  m)
(C) acceleration of man is
Mg
(M  m)
(D) measured mass of man is M.
Ans: (A, C)
Mg – T = Ma
T = ma
…(i)
…(ii)
16
Solving (i) and (ii)
a
Mg
(M  m)
FBD of man
N
a
M g
Mg – N = Ma
N
Mmg
(M  m)
18. Figure shows two blocks A and B connected to an ideal
pulley string system. In this system when bodies are
released then (neglect friction and take g = 10 m/s2)
10 kg
B
40 kg
A
a
(A) Acceleration of block A is 1 m/s2
(B) Acceleration of block A is 2 m/s2
(C) Tension in string connected to block B is 40 N
(D) Tension in string connected to block B is 80 N
Ans: (B, D)
17
10 kg
A
Applying NLM on 40 kg block 400 – 4T = 40 a
For 10 kg block T = 10.4 a
Solving a = 2 m/s2
T = 80 N
19. The system of two blocks is at rest as shown in the figure.
A variable horizontal force is applied on the upper block.
The co–efficient of friction for both the contacts is .
Choose the correct option(s)
m

F
G
ro
u
n
d
2
m
(A) When acceleration of the upper block is 2 g net force
on the lower block by the ground in mg 9   .
(B) When acceleration of the upper block is 2 g
acceleration of the lower block is zero.
(C) Net horizontal force on the lower block is always zero.
(D) Unless the upper block moves, no frictional force exists
between the ground and the lower block.
Ans: (A, B, C)
2
f
1
m F
(
f
)
=

m
g
1
m
a
x
f
2
2
m
f
1
(
f
)
=
3

m
g
2
m
a
x
18
(f1)max = mg < (f2)max
 2m will not move.
When acceleration of m is 2g then
F = 3 mg and f1 = mg.
 Total force on (2m) by ground  f
2
2
 (3mg)2  mg 9   2
.
20. The figure shows a block of mass m placed on a smooth
wedge of mass M. Calculate the value of M and tension in
the string, so that the block of mass m will move vertically
downward with acceleration 10 m/s2 (Take g = 10 m/s2)
m


Smooth


(A) the value of M is
(B) the value of M is
M cot 
1  cot 
M tan 
1  tan 
(C) the value of tension in the string is
(D) the value of tension is
Mg
tan 
g
cot 
Ans: (A, C)
R
T
M
Mg
M 'g  T  M 'a
T = Ma
…(i)
…(ii)
19
M 'g  a(M  M ')
a
M 'g
(M  M ')
ma sin   mg cos 
a  g cot 
g cot  
M 'g
(M  M ')
masin
ma

mgcos
mgsin
+ macos
cot  M  cot M '  M '
M' 
M cot 
(1  cot )
T=Ma
= M. g cot 
Mg
.
T
tan 
Numeric Type
21. Rain is falling vertically with a speed of 20 m/s. A person
is running in the rain with a velocity of 5 m/s and a wind is
also blowing with a speed of 15 m/s both from the west.
The angle with the vertical at which the person should hold
his umbrella so that he may not get drenched is tan-1(n/2). n
=?
Ans: (1)
20
j
i
E
D o w n w a rd
Vr /p  Vr  Vp  20ˆj  15iˆ  5iˆ  20ˆj  10iˆ
1
   tan 1   .
2
22. In the arrangement shown in figure, coefficient of friction
between the two blocks is   1/ 2. Find the force of friction
acting between the two blocks.
Ans: (8)
Free body diagram of the two bodies are as follows
Let acceleration of both the blocks towards left is a. Then
a
f  2 20  f

2
4
or 2f  4  20  f
or f  8N
Maximum friction between the two blocks can be:
f max  mg
=(0. 5) (2) (10) = 10 N
Now since f  f max
Therefore, friction force between the two blocks is 8 N.
21
23. At certain moment of time, velocities of 1 and 2 both are 1
m/s upwards. The velocity of 3 at that moment is x m/s
downward. What is the value of x?
1
2
3
Ans: (3)
1
2
3
a2  a3  2a1  0
Similarly, we can find
v2  v3  2v1  0
Taking, upward direction as positive we are given
v1  v2  1 m/s
 v3  3 m/s
i.e., velocity of blocks 3 is 3 m/s (downwards).
24. Block B has a mass m and is released from rest when it is
on top of wedge A, which has a mass 3m. The tension in
22
cord CD while B is sliding down A is
 mg

sin 2  .

 n

Neglect
friction. Value of n is?
B
DC
mA
 3
Ans: (2)
Normal reaction between A and B would be N  mg cos  . Its
horizontal component is N sin . Therefore tension is cord
CD is equal to this horizontal component.
Hence, T  N sin   (mg cos )(sin )  mg sin 2
2
25. Three blocks A, B and C of masses 3 kg, 2 kg and 1 kg
respectively connected by two ideal strings are present on a
smooth horizontal surface. An external horizontal force of
18 N acts on the block A to pull the system. Find the
difference in the tensions in strings connecting A & B and,
B & C (in Newton).
Ans: (6)
A B C
T
T
2
1
3
1
2
F
a
18
 3m / s 2
6
F – T1 = 3a
T1 = 9N
T2 = 1a
 T2 = 3 N
T1 – T2 = 6 N.
23
F
1
26.
2
k
g
F
2
1
k
g
s
m
o
o
th
F1  12N & F2  3N
find action – reaction force between the two
objects
Ans: (6)
1
2
N
3
N
R
R
1
2
12 – R = 2a
R – 3 = 1a
By (i) and (ii)
a = 3 m/s2
R = 6 N.
…(i)
…(ii)
27. In figure if  = 0.1 is the coefficient of friction between the
wedge and the horizontal surface then for what maximum
value of m in kg, the wedge remains at rest [take M = 18 kg
and  = 45º].
sm
o
o
th
m
M


Ans: (4)
N
a
mgsin
mgcos
F.B.D of m
N = mg cos  ….(i)
mg sin  = ma
a = g sin 
24
F.B.D. of M
R
N
co
s
N
sin


R
M
g
R = N cos  + Mg
 R = N sin 
. (N cos  + Mg) = N sin 
 Mg = N [sin  -  cos ]
 Mg = mg cos  [sin  -  cos ]
m
M
cos [sin    cos ]
m = 4 kg.
28. Find the mass M (in kg) of the hanging block in figure
which will prevent the smaller block from slipping over the
triangular block. All the surfaces are frictionless and the
strings and the pulleys are light. Take  = 37o, M '  1 kg ,
1
m  kg .
3
m
M

M
Ans: (4)
For m to be at rest with respect to M ' , a = g tan 
 T = (M+ m) g tan 
For block M, Mg   M ' m  g tan   Mg tan 
25
On solving,
M
M ' m
4
cot   1
29. Figure shows a small block A of mass m kept at the left end
of a plank B of mass M = 2 m and length l. The system is
started moving towards right with the initial velocity v and
placed on the horizontal surface. The friction coefficients
between the road and the plank is 1/2 and that between the
plank and the block is 1/4. If the time elapsed before the
block separates from the plank is
n
l
,
3g
then value of n is?
A
B
l
Ans: (4)
There will be relative motion between block and plank and
plank and road. So at each surface limiting friction will act.
The direction of friction forces at different surfaces are as
shown in figure.
A
f1
B
f1
f2
Here,
1
f1    (mg)
4
1
3
f 2    (m  2m) g    mg
2
2
Retardation of A is a1  f1  g
m 4
and retardation of B is a2  f 2  f1  5 g
2m
8
and
26
Relative acceleration of A with respect to B is
ar  a2  a1 
3
g
8
Initial velocity of both A and B is v, so, there is no relative
initial velocity.
Hence, applying s  1 at or l  1 a t  3 gt
2
2
2
2
t  4
r
2
16
l
3g
30. In the fig. the distance BQ=3 m, BP = 14 m at time t = 0.
The system of blocks is released from rest at time t = 0.
The string connecting B and C is suddenly cut at time
t = 2s. Calculate the velocities (m/s) of A and B at the
instant when the block B hits the pulley Q. the coefficient
of friction between B and the horizontal surface is
2
    0.25. Take g = 9.8 m/s
s
k
4
k
g
P
B
C8
k
g
Q
2
k
gA
Ans: (7)
Initially 8 g – T – 8a
1
T  T   4g  4a
4
T  2g  2a
Solving, we get
a a  5g left words.
B
Now,
14
T0
at
t  2s v B 
5g
5g
, sB 
7
7
27
2g  T0  2a 
g  T0  4a 
2g  6a  a  
g
2
58 g
 t0
7 2
t 5
10
 t s
2 7
7
v
2
g
 58 
0     2 t  0
2
 7 
25g
s
m
4g
Now, f gets reserved.
2g  T0  2a
T0  TK  4a
g
6
g  25g 5g

v 2  02  2  

 3
6  49
7

g  ta  a 
g
v 2  15  v 2  5  9.8
3
 49.0  v  7ms 1
28