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 \(n=\)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(24\)
Step-by-step Solution
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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