JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(Nm ^{-1}\).
- A \(680\)
- B \(700\)
- C \(608\)
- D \(600\)
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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