ExamBro
ExamBro
MHT CET · Physics · Work Power Energy

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)

  1. A \(\frac{2 \mathrm{mg}(\mathrm{s}-\mathrm{h})}{\mathrm{h}^2}\)
  2. B \(\frac{2 \mathrm{mg}(\mathrm{s}+\mathrm{h})}{\mathrm{h}^2}\)
  3. C \(\frac{\mathrm{mg}(\mathrm{s}-\mathrm{h})}{\mathrm{h}}\)
  4. D \(\frac{\mathrm{mg}(\mathrm{s}+\mathrm{h})}{\mathrm{h}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{2 \mathrm{mg}(\mathrm{s}+\mathrm{h})}{\mathrm{h}^2}\)

Step-by-step Solution

Detailed explanation

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,
\(\begin{aligned}
& \frac{1}{2} \mathrm{Kh}^2=\mathrm{mg}(\mathrm{S}+\mathrm{h}) \\
\therefore \quad & \mathrm{K}=\frac{2 \mathrm{mg}(\mathrm{S}+\mathrm{h})}{\mathrm{h}^2}
\end{aligned}\)