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

A \(0.5 \,kg\) block moving at a speed of \(12 \,ms ^{-1}\) compresses a spring through a distance \(30\, cm\) when its speed is halved. The spring constant of the spring will be \(Nm ^{-1}\).

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Answer & Solution

Correct Answer

(D) \(600\)

Step-by-step Solution

Detailed explanation

\(U _{ i }+ K _{ i }= U _{ f }+ K _{ f }\)

\(\Rightarrow 0+\frac{1}{2} m (12)^{2}=\frac{1}{2} K (0.3)^{2}+\frac{1}{2} m (6)^{2}\)

\(\Rightarrow 0.5\left(12^{2}-6^{2}\right)= K (0.3)^{2}\)

\(K =600 \,N / m\)

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