Question

Seema tried to push a heavy rock of 100 kg for 200s but could not move it. Find the work done by Seema at the end of 200s.

Answer 1

Hint: Work is actually dependent on displacement, the work done by a certain force for moving a body to a certain distance is actually proportional to the distance covered or more appropriately the displacement occurred.

Formula Used:
$W = F\times s$
Where $W$ is the work done, $F$ is the force and $s$ is the displacement.

Complete step by step answer:
Let F be the force applied by Seema and s be the displacement in 200s
Given in the question, that Seema cannot move the block. Thus we can conclude that,
$ \Rightarrow s = 0$
We know that,
$W = F\times s$
Substituting the value of $s$ in the above equation we get
$ \Rightarrow W = F\times0$
$ \Rightarrow W = 0J$
Thus work done is $0J$

Additional Information: Work done is the product of force and displacement in the direction of force. It is measured in Joules (represented as J). It is denoted by $W$. Dimension of work done is $[{M^1}{L^2}{T^{ - 2}}]$
The formula for work is
$W = \vec F \cdot d\vec s$
Work done is a scalar quantity since it is the scalar product of force and displacement. Thus it has no direction. Displacement is the shortest distance between two points. Its S.I unit is meter (m). It can be positive, negative or zero depending on the situation. The magnitude of displacement is always less than or equal to distance. Force is defined as a pull or pull. It is mathematically the product of mass and acceleration. It is denoted by $F$ and has unit $N$ (Newton). The dimension of Force is $[ML{T^{ - 2}}]$

Note: Take care that time is given just for confusion. No need for time in this question. Also the value of displacement will be 0 in this case. Do not forget to put this in the equation of work done. Reading and understanding the question properly is the key to solve this question.