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