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:
Show 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 $mg$.
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:
\[
W = N \cdot s = \frac{3mg}{2} \times \frac{gt^2}{4}
= \frac{3mg^2 t^2}{8}
\]
