Question

In which of the following cases work is said to be done?
A.A man pushing a roller and displacing it
B.A boy sleeping
C.Girl writing in exam
D.All of these

Answer 1

Work is when energy is transferred into a body via application of external force on it and that energy causes a change in the position of the body in the direction of application of force.

Complete answer:
Work is done only when a body moves from one position to another and that too, due to the action of some external force. The external force which causes the displacement is said to have done the work.
For unidirectional motion, the value of the work done can be expressed as a multiplication product of the force applied and the displacement achieved by the application of that force provided that the displacement is in the direction of the force applied.
For two directional or three directional motion, we need to consider the direction vectors involved. We know that force as well as displacement are both vector quantities which means that they have both direction as well as magnitude. And the interesting thing is work is a scalar quantity. Thus to find out the work done, we use the scaler (dot) product of force and displacement.
$\ W = \overrightarrow F \bullet \overrightarrow S \\\ = \left( {FS} \right)\hat f \bullet \hat s \\\ \$
Where F and S are the magnitudes of the force and displacement whereas $\hat f$ and $\hat s$ are the direction vectors of the same.
In the above question, let's analyse the options. The man in the first option causes displacement of the roller by applying force so we can say that he does work.
The boy in the second option is sleeping and although he applies gravitational force on the bed, the displacement does not occur. As such there is no work done in this case.
The girl in the third option is writing and so she applies muscular as well as gravitational force on the paper but if displacement occurs, she will naturally not be able to write and if she is writing, this signifies that the paper is stationary. As such there is no work done in this case too.
So work is done only in the first case.

Note:
If the displacement is constantly changing, then we use integration to find out the work done and sections of the displacement motion are observed. $w = \int {F.ds}$ where w is work, F is force and ds is the section of motion being observed at any particular instance.