Question

A block of mass $\text{'}\mathrm{m}\text{'}$ is kept on a platform which starts from rest with constant acceleration $\frac{\mathrm{g}}{2}$ upward, as shown in fig. Work done by normal reaction on block in time $\text{'}\mathrm{t}\text{'}$ is :

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

Answer 1

The correct option is A

$\frac{3{\mathrm{mg}}^2{\mathrm{t}}^2}{8}$

Step 1: Given data and drawing the diagram

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{block}=\mathrm{m} \\ \mathrm{Upward}\mathrm{acceleration},a=\frac{\mathrm{g}}{2} \\ \mathrm{Time}=\mathrm{t} \\ \mathrm{Initial}\mathrm{velocity},\mathrm{u}=0 \end{gathered}$$

The diagram showing all the forces acting on the block is given below:

Here we see that normal force, $\text{'}\mathrm{N}\text{'}$and $\text{'}\mathrm{mg}\text{'}$ are in opposite direction. Also there is an acceleration in upward direction, hence there will be a pseudo force acting on the block which is given below:

${\mathrm{F}}_{\mathrm{p}}=\mathrm{m}\frac{\mathrm{g}}{2}$

Step 2: Calculating the work done by normal reaction

Apply the force balance on the given block we get,

$$\begin{gathered} N-mg=m\frac{g}{2} \\ \Rightarrow N=\frac{3mg}{2} \end{gathered}$$

Also the displacement can be calculated by using equation of motion.

$$\begin{gathered} \mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{S}=0+\frac{1}{2}\left(\frac{\mathrm{g}}{2}\right){\mathrm{t}}^2(\because \mathrm{u}=0\mathrm{and}\mathrm{a}=\frac{\mathrm{g}}{2}) \\ \mathrm{S}=\frac{{\mathrm{gt}}^2}{4} \end{gathered}$$

Now work done can be calculated using the formula:

$$\begin{gathered} \mathrm{W}=\mathrm{NS} \\ \mathrm{W}=\left(\frac{3\mathrm{mg}}{2}\right)\left(\frac{{\mathrm{gt}}^2}{4}\right) \\ \mathrm{W}=\frac{3{\mathrm{mg}}^2{\mathrm{t}}^2}{8} \end{gathered}$$

Hence, option(A) is correct answer.