WORK-ENERGY-POWER
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(i) WORK:-
Work done by a force is the product of the force and
the distance moved by the point of application in the
direction of the force. It is a scalar quantity.
A
F
α
B
s
F
α
Unit: Nm ( Joule )
Work done=F × cosα × s
α= angle of inclination of force with the
direction of motion
s=displacement of force from A to B 1
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(ii) POWER:It is defined as the time rate of doing work.
Power = work done /Time= (force × distance) /Time
= force × velocity
Unit: (Nm)/s [watt]
(kN m)/s [kilo watt]
1 metric H.P=735.75 watts.
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iii) Energy:It is defined as the capacity to do work. It is a scalar
quantity.
Unit :- N m (Joule)
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Work-Energy relation for translation
From Newton’s second law of motion
∑ F =(W/g) × a -----------(1)
Also
a = dv/dt =( dv/ds) ×( ds/dt) = v × dv/ds
sub in (1)
∑ F=( W/g ) × v × dv/ds
F × ds = (W/g ) × v × dv ------------------(2)
Let the initial velocity be u and the final velocity after it
moves through a distance ‘ s’ be v
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Integrating both sides, we get
s
v
 F  ds = (W/g)  v dv
0
u
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V
 F × s = ( W/g ) v2/2 ]
u
∑ F × s = (W/2g) [v2-u2]
Therefore work done by a system of forces acting on a body
while causing a displacement is equal to the change in
kinetic energy of the body during the displacement.
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Work done by a spring
s
F=k(s)
The force required to cause unit deformation of the spring is called the
spring modulus denoted by the symbol k. The force required to deform
a spring is given by F= ks. Work done by the force on a spring is the
product of the average force and the deformation s.
W= -ks2/2. [ The negative sign indicates that whenever spring is
deformed the force of spring is in the opposite direction of deformation.]
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IMPULSE – MOMENTUM
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Momentum:Quantity of motion possessed by a body is called
momentum. It is the product of mass and velocity. It is a
vector quantity.
Unit:- N s.
Impulse of a Force:It is defined as the product of force and the time over which
it acts. It is a vector quantity.
Unit:- N s.
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Impulse-momentum relationship
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Force = Rate of change of momentum
F =( mv – mu) /t = Force causing impulse
F×t = mv – mu
Impulse = final momentum – Initial momentum
The component of the resultant linear impulse along
any direction is equal to change in the component of
momentum in that direction.
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Law of Conservation of
momentum
The law of conservation of momentum may be
stated as , “momentum is conserved in a system in
which resultant force is zero”. In other words, in a
system if the resultant is zero, Initial momentum is equal
to Final momentum
m1u1+m2u2 = m1v1+m2v2
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ASSIGNMENT
1.A small block starts from rest at point a and slides down the inclined
plane. At what distance along the horizontal will it travel before coming
to rest . Take µk=0.3 [Ans :s=6m ]
A
3
4
C
B
s
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2.The system starts from rest in the position shown . How much
further will block ‘A’ move up the incline after block B hits the
ground . assume the pulley to be frictionless and massless and µ is 0.2
.WA=1000N, WB=2000N. [ Answer s =1.27m]
A
3
4
B
3m
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3.Two bodies a and b weighing 2000N and 5200N are connected
as shown in the figure . find the further distance moved by block a
after the block b hits block c .[ Answer s=1.34m]
A
5
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B
4.A 1500Kg automobile travels at a uniform rate of 50kmph to 75kmph .
During the entire motion, the automobile is traveling on a level
horizontal road and rolling resistance is 2 % of weight of automobile .
Find (i) maximum power developed (ii) power required to maintain a
constant speed of 75kmph.[ ANSWER: power developed = 6.131KN]
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5.A spring is used to stop 60kg pack age which is sliding on a
horizontal surface . the spring has a constant k which is 20kN/m and
is held by cable such that it is initially compressed at 120mm.
knowing that the package has a velocity of 2.5m/s in position shown
and maximum additional displacement of spring is 40mm .
Determine the co-eff of kinetic friction between package and surface.
(Answer µk=0.2)
2.5 m/s
60kg
600m
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6. The system shown in figure has a rightward velocity of 4m/s, just
before force P is applied. Determine the value of P that will give a
leftward velocity of 6m/s in a time interval of 20sec. Take µ = 0.2 &
assume ideal pulley. [Answer P=645.41N]
P
1000N
400N
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7. A locomotive of weight 500kN pulls a train of weight of 2500kN.
The tractive resistance, due to friction is 10N/kN. The train can go with
a maximum speed of 27kmph on a grade of 1in100. Determine (a)
Power of the locomotive. (b) Maximum speed it can attain on a straight
level track with the tractive resistance remaining same. [Answer (a)
Power= 450kN (b) v=15m/s]
8.A wagon weighing 400kN starts from rest, runs 30m down a 1%
grade & strikes a post. If the rolling resistance of the track is 5N/kN,
find the velocity of the wagon when it strikes the post.
If the impact is to be cushioned by means of one bumper string, which
compresses 1mm per 20 kN weight, determine how much the bumper
spring will be compressed. [Answer v=1.4m/s, x=63.21mm]
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9. A car weighing 11,100 N & running at 10m/s holds 3 men each
weighing 700N. The men jumps off from the back end gaining a relative
velocity of 5m/s with the car if the three men jumps off (i) in succession,
(ii) all together. [Answer (i) v=10.85m/s. (ii) v=10.8m/s.]
10. A gun weighing 300kN fires a 5kN projectile with a velocity of
300m/s. With what velocity will the gun recoil? If the recoil velocity
is over come by an average force of 600N, how far will the gun travel
& how long will it travel. [Answer v=-5m/s s=0.637m/s t=0.255s ]
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