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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2022
  • Hard
  • Numerical

A block of mass ' \(m\) ' (as shown in figure) moving with kinetic energy \(E\) compresses a spring through a distance \(25\,cm\) when, its speed is halved. The value of spring constant of used spring will be \(nE\; Nm ^{-1}\) for \(n=\)

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Answer & Solution

Correct Answer

(D) \(24\)

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Detailed explanation

Using work - energy theorem

\(W _{\text {net }}=\left( K _{ f }- K _{ i }\right)\)

\(\Rightarrow-\frac{1}{2} Kx ^{2}=\frac{1}{2} m \left(\frac{ v }{2}\right)^{2}-\frac{1}{2} mv ^{2}=\frac{ E }{4}- E\)

\(\Rightarrow \frac{1}{2} Kx ^{2}=\frac{3 E }{4} \Rightarrow K =\frac{3 E }{2 x ^{2}}\)

\(\Rightarrow K =\frac{3 E }{2 \times\left(\frac{1}{4}\right)^{2}}=24\,E\)

\(n =24\) 

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