Question

In which of the following work is being done?
A. The man sitting on the bench
B. A person standing with a basket of fruit on the head
C. Climbing the tree to pluck
D. Pushing a Wheelbarrow of bricks

Answer 1

This question is based on the work done formula. You need to calculate the work done by the formula $W = F \times d\cos \theta$, Where $\theta$ is the angle between the force's line of action and the displacement vector d. You need to calculate the work done in each case by using the work formula.

Complete step by step answer:
Case-1 The man sitting on a bench
In this case, the weight of the man is acting downwards, and as the man is sitting on the bench, then there is no movement of the man, so the displacement is zero$\left( {d = 0} \right)$ . Hence, according to the work done expression,
$\begin{array}{l} W = \left( {weight\;of\;the\;man} \right) \times d\cos \theta \\\ \Rightarrow W = \left( {weight\;of\;the\;man} \right) \times 0 \times \cos \theta \\\ \Rightarrow W = 0 \end{array}$
Hemce, the work done is zero.

Case-2 Person standing with a basket of fruit on the head
The person is standing with the basket of fruit on the head , so the weight of the basket is acting downwards and also the person is not moving, so there is no displacement$\left( {d = 0} \right)$,
Hence, according to the work done expression,
$\begin{array}{l} W = \left( {weight\;of\;the\;fruit\;basket} \right) \times d\cos \theta \\\ \Rightarrow W = \left( {weight\;of\;the\;fruit\;basket} \right) \times 0 \times \cos \theta \\\ \Rightarrow W = 0 \end{array}$
Thus, work done in this case is also zero.

Case-3 Climbing the tree to pluck.
Climbing tree means the man has changed its location, so there is a displacement $\left( {d \ne 0} \right)$ and the direction of displacement and the weight of the man is in the same direction so $\theta = {0^\circ }$, that means there is work done.
Hence, according to the work done expression,
$\begin{array}{l} W = \left( {weight\;of\;the\;man} \right) \times d\cos 0\\\ \Rightarrow W = \left( {weight\;of\;the\;man} \right) \times d \times 1\\\ \Rightarrow W \ne 0 \end{array}$

Case-4 Pushing a Wheelbarrow of bricks
Pushing the wheelbarrow of bricks means that the wheelbarrow of bricks is displaced from one position to another position, so there is displacement, and the line of action of force and the displacement’s direction is the same so, $\theta = {0^\circ }$. Then the work done in this case will not be zero.
Hence, according to the work done expression,
$\begin{array}{l} W = \left( {push\;force} \right) \times d\cos 0\\\ \Rightarrow W = \left( {push\;force} \right) \times d \times 1\\\ \Rightarrow W \ne 0 \end{array}$

So, the correct answer is “Option C and D”.

Note:
You may go wrong in calculating the angle $\theta$ here. There is $\cos \theta$ in the formula then, you may do mistake in calculating the final work done because when$\theta = {0^\circ }$then $\cos \theta = 1$and when$\theta = {90^\circ }$then $\cos \theta = 0$.