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WORKPOWER
ENERGY
, QUIZRR 3
WORK, ENERGY & POWER
1
1 .1 . WORK
It refers to an activity involving a force and movement in the direction of the force.
1 .1 .1 Wo r k d o n e b y a c o n s t a n t f o r c e
When the point at which a force acts moves, the force is said to have done work.
When the force is constant, the work done is defined as the product of the force and distance moved.
Work done = force distance moved in direction of force
Consider in Figure a force F acting at the angle moves a body from point A to point B.
F
os
sc
A B
s
The distance moved in the direction of the force is given by
Distance in direction of force = s cos
So the work done by the force F is
Work done = F s cos
If the body moves in the same direction as the force the angle is 0.0 so
Work done = Fs
When the angle is 90 then the work done is zero.
The SI units for work are Joules J (with force F, in NewtonÊs N and distance, s, in metres m).
Note : Under the action of a constant force work done is path independent,
i.e., it depends on initial and final positions only. For example, if an object takes 1 B
from position A to position B through three different paths 1, 2 and 3 under a
2
constant force say F 2i→ 3 j k→ , then work done by this force in all three
paths will be same, or A
3
W1 = W2 = W3
Fig. 4.4
Similarly, work done by force of gravity m g near the surface of earth is path independent as
force of gravity is constant. Although we will see later that being a conservative force work done by force
of gravity is path independent at greater heights also.
Now let us take few examples related to above article.
WORK, ENERGY & POWER
, 4 QUIZRR
Ex a m p l e 1
How much work is done when a force of 5 kN moves its point of application 600 mm in the
direction of force.
Solution : Work done = (5 103) (600 10 3)
= 3000J
= 3 kJ
Ex a m p l e 2
Find the work done in raising 100 kg of water through a vertical distance of 3 m.
Solution : The force is the weight of the water, so
work done = (100 9.81) 3
= 2943 J
Ex a m p l e 3
Two unequal masses of 1 kg and 2 kg are attached at the two ends of a light
inextensible string passing over a smooth pulley as shown in figure. If the
system is released from rest, find the work done by string on both the blocks
in 1 s. Take g = 10 m/s2.
1kg
2kg
Solution : Net pulling force on the system is Fig. 4.6
Fnet = 2g 1g = 20 10 = 10 N
Total mass being pulled m = (1 + 2) = 3 kg
Therefore, acceleration of the system will be
Fnet 10
a m/s2
m 3
Displacement of both the blocks in 1 s is
1 1 10 2 5
S at2 1 m
2 2 3 3 a T
1kg
Free body diagram of 2 kg block is shown in figure.
a
Using F = ma, we get
2kg a
10 1g 2kg
20 T 2a 2
3
2g 20 N
20 40
or T 20 N
3 3
Work done by string (tension) on 1 kg block in 1 s is
W1 = (T) (S) cos 0
40 5 200
3 3 1 9 J Ans.
WORK, ENERGY & POWER
, QUIZRR 5
Similarly, work done by string on 2 kg block in 1 s will be
W1 = (T) (S) (cos 180 )
40 5 200
3 3 1 9 J Ans.
1 .1 .2 Wo r k d o n e b y a v a r ia b l e f o r c e
Forces in practice will often vary. Consider the case where the force varies as in Figure.
For the thin strip with width ds - shown shaded in Figure the force can be considered constant at
F. The work done over the distance ds is then
work done = F ds
This is the area of the shaded strip.
The total work done for distance s is the sum of the areas of all such strips. This is the same as the
area under the Force-distance.
force
F
0 distance
ds
Work done = area under force/distance curve
Clearly, this also works for a constant force the curve is then a horizontal line.
1 .1 .3 Ge n e r a l Co n c e p t
If point of application of force suffers an infinitesimal vector d r relative to a frame of reference S, then
infinitesimal work done by force during infinitesimal displacement, relative to same frame of reference S.
d W F.d r
dr F
Two important points to be noted,
r
(i) frame of reference must be mentioned
(ii) point of application is important in case of rigid body.
W dW F.d r
WORK, ENERGY & POWER
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