Q131 / 150 · MHT CET 2023 (10 May Shift 2) · Physics
A ball of mass ' $\mathrm{m}$ ' is dropped from a height ' $\mathrm{s}$ ' on a horizontal platform fixed at the top of a vertical spring. The platform is depressed by a distance ' $h$ '. The spring constant is ( $\mathrm{g}=$ acceleration due to gravity)
- A$\frac{2 \mathrm{mg}(\mathrm{s}-\mathrm{h})}{\mathrm{h}^2}$
- Correct answerB$\frac{2 \mathrm{mg}(\mathrm{s}+\mathrm{h})}{\mathrm{h}^2}$
- C$\frac{\mathrm{mg}(\mathrm{s}-\mathrm{h})}{\mathrm{h}}$
- D$\frac{\mathrm{mg}(\mathrm{s}+\mathrm{h})}{\mathrm{h}}$
Solution
From the question, it can be understood that the total distance the ball falls is $(\mathrm{S}+\mathrm{h})$
The spring is compressed through a length $h$
$\therefore \quad$ Loss of P.E. by the ball $=\mathrm{mg}(\mathrm{S}+\mathrm{h})$
Work done on the spring $=\frac{1}{2} \mathrm{Kh}^2$
Using law of conservation of energy,
$\begin{aligned}
& \frac{1}{2} \mathrm{Kh}^2=\mathrm{mg}(\mathrm{S}+\mathrm{h}) \\
\therefore \quad & \mathrm{K}=\frac{2 \mathrm{mg}(\mathrm{S}+\mathrm{h})}{\mathrm{h}^2}
\end{aligned}$
Hence, the correct answer is option: 'B'.