Question

A man of mass $50kg$ is standing in an elevator. If the elevator is moving up with an acceleration $\frac{g}{3}$ then what is the work one by normal reaction of the floor of the elevator on the man when elevator moves by a distance 12m? $\left(g=10m/s^2\right)$

• A. 8000 J
• B. 6000 J
• C. 4000 J
• D. 2000 J

Answer 1

The correct option is A

8000 J

Step 1: Given

Mass of the man: $m=50kg$

Distance moved by lift: $d=12m$

Acceleration due to gravity is $g$.

Acceleration of the lift: $a_L=-\frac{g}{3}$ (negative because the lift is going upwards, moving against earth's gravity)

The resultant acceleration is $a$.

Work done is $W$.

Step 2: Formula used

• $F=m\times a$
• $W=F\times d$

Step 3: Find the resultant acceleration by subtracting the acceleration of lift from acceleration due to gravity.

$$\begin{gathered} a=a_L-g \\ =-\frac{g}{3}-g \\ =-\frac{g}{3}-g \\ =-\frac{4g}{3} \end{gathered}$$

Step 4: Find the force using the formula and substitute the value of g and m (Take only the magnitude of acceleration as the negative sign only indicate direction and have no affect on value).

$$\begin{gathered} F=m\times a \\ =50kg\times \frac{4g}{3}m/s^2 \\ =50kg\times \frac{4\times 10}{3}m/s^2 \\ =\frac{2000}{3}N \end{gathered}$$

Step 5: Find the work done using the formula

$$\begin{gathered} W=F\times d \\ =\frac{2000}{3}N\times 12m \\ =8000J \end{gathered}$$

Hence, option (A) is correct.