Question

A man of mass $50kg$ is standing in an elevator. If elevator is moving up with an acceleration $\dfrac{g}{3}$ then work done by normal reaction of floor of elevator on man when elevator moves by a distance $12m$ is $\left( {g = 10\dfrac{m}{{{s^2}}}} \right)$.
$(A)2000J$
$(B)4000J$
$(C)6000J$
$(D)8000J$

Answer 1

Hint: This question is based on Newton's third law of motion which states that for every action, there is an equal and opposite reaction. The expression for the Newton’s third law of motion, which is also known as action-reaction law, is:
${v^2} - {u^2} = 2as$

Complete step by step answer:
The work which is done on the man by the floor will be come out to be equal to the sum of change in potential energy and kinetic energy,
$W = \Delta PE + \Delta KE$
$W = mgh + \dfrac{1}{2}m{v^2}$
This can be re-written as,
$W = m\left( {gh + \dfrac{{{v^2}}}{2}} \right)......(1)$
In order to solve this, we have to calculate the value of $v$
Using the third equation of motion:
$v_f^2 - v_i^2 = 2as$
We know that the elevator starts from rest.
So, the velocity after travelling $12m$ is,
${v^2} - 0 = 2 \times \dfrac{{10}}{3} \times 12$
On solving this, we get,
${v^2} = 80$
On putting the above value in equation (1), we get,
$W = 50\left( {10 \times 12 + \dfrac{{80}}{2}} \right)$
$W = 50 \times 160$
$W = 8000J$
So, the correct answer is $(D)8000J$.

Note: It is important to that in this particular question, the value of acceleration due to gravity is given as $\left( {g = 10\dfrac{m}{{{s^2}}}} \right)$. The actual value of acceleration due to gravity is $\left( {g = 9.8\dfrac{m}{{{s^2}}}} \right)$ but in this question it is approximated to $\left( {g = 10\dfrac{m}{{{s^2}}}} \right)$, in order to make the calculation part easy.