Answer
427.5k+ views
Hint: The answer to both the questions lies in the definition of work done, which is a scalar quantity. The work done by a force or against a force is the dot product of force vector and displacement vector.
Formula used:
Work done is a scalar quantity given by:
$W = \vec{F}. \vec{s}$
Complete answer:
(a) In the first case, a weight lifter has been given to be holding a weight of 120 kg on his shoulder for 30 s. The displacement of the weight for the duration of 30 s is practically zero. So, even if it seems like a tough job, there is no work done on the weight for the 30 s duration.
Therefore, the work done is zero in this case as s is zero.
(b) In calculation of the work done by a locomotive in carrying the weight over a plane road we have to take into account the direction of force and direction of displacement. The force here is the force of gravity acting downwards and as the locomotive is travelling on a plane path, the displacement vector happens to be in a direction which is perpendicular to the direction of force. As we know
$W = \vec{F}. \vec{s} = Fs \cos \theta$,
and as
$W = Fs \cos 90^{\circ} = 0$.
Therefore, in this case too, work done is zero.
Additional information:
There is an important concept of conservative force that determines the nature of work done. If a force is conservative, then the amount of work done depends on the endpoints only and not on the path travelled.
The force of gravity is a conservative force. Therefore as we lift a mass at some height for 30 s when the amount of work done will be the difference in potential energy at t = 30 and t = 0 time. In between, anything could have happened; it will not affect our answer. The point here is before and after 30 seconds the mass stays at the same height so no work is done.
Note:
The formation of the question itself gives the hint for the answer. For the first case, even if one thinks of applying the formula for potential energy of the weight at a height, we are not given the height and in the second part too, magnitude of displacement and force is not given. Therefore, the correct answer obviously seems to be zero in both cases else the data would be insufficient.
Formula used:
Work done is a scalar quantity given by:
$W = \vec{F}. \vec{s}$
Complete answer:
(a) In the first case, a weight lifter has been given to be holding a weight of 120 kg on his shoulder for 30 s. The displacement of the weight for the duration of 30 s is practically zero. So, even if it seems like a tough job, there is no work done on the weight for the 30 s duration.
Therefore, the work done is zero in this case as s is zero.
(b) In calculation of the work done by a locomotive in carrying the weight over a plane road we have to take into account the direction of force and direction of displacement. The force here is the force of gravity acting downwards and as the locomotive is travelling on a plane path, the displacement vector happens to be in a direction which is perpendicular to the direction of force. As we know
$W = \vec{F}. \vec{s} = Fs \cos \theta$,
and as
$W = Fs \cos 90^{\circ} = 0$.
Therefore, in this case too, work done is zero.
Additional information:
There is an important concept of conservative force that determines the nature of work done. If a force is conservative, then the amount of work done depends on the endpoints only and not on the path travelled.
The force of gravity is a conservative force. Therefore as we lift a mass at some height for 30 s when the amount of work done will be the difference in potential energy at t = 30 and t = 0 time. In between, anything could have happened; it will not affect our answer. The point here is before and after 30 seconds the mass stays at the same height so no work is done.
Note:
The formation of the question itself gives the hint for the answer. For the first case, even if one thinks of applying the formula for potential energy of the weight at a height, we are not given the height and in the second part too, magnitude of displacement and force is not given. Therefore, the correct answer obviously seems to be zero in both cases else the data would be insufficient.
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