Question

A block of mass $m$ is kept on a platform which starts from rest with constant acceleration $\dfrac{g}{2}$ upward, as shown in the figure. Work done by the normal reaction on the block in time $t$ is:

• A. $\dfrac{mg^2 t^2}{8}$
• B. $0$
• C. $-\dfrac{mg^2 t^2}{8}$
• D. $\dfrac{3mg^2 t^2}{8}$

Hint:

Work done by a force equals force $\times$ displacement in the direction of the force. When a platform accelerates upward, the normal reaction is greater than $mg$.

Answer 1

The Correct Option is D

Step 1: Identify the forces acting on the block. Upward force = Normal reaction $N$ Downward force = Weight $mg$
Step 2: Apply Newton’s second law to the block (upward positive): \[ N - mg = m\left(\frac{g}{2}\right) \]
Step 3: Solve for the normal reaction: \[ N = mg + \frac{mg}{2} = \frac{3mg}{2} \]
Step 4: Find the displacement of the block in time $t$. Since the platform starts from rest with acceleration $\frac{g}{2}$: \[ s = \frac{1}{2}at^2 = \frac{1}{2}\left(\frac{g}{2}\right)t^2 = \frac{gt^2}{4} \]
Step 5: Calculate the work done by the normal reaction: \[ W = N \cdot s = \frac{3mg}{2} \times \frac{gt^2}{4} = \frac{3mg^2 t^2}{8} \]