5.Work, Energy, Power and Collision
medium

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

$600$

(JEE MAIN-2022)

Solution

$U _{ i }+ K _{ i }= U _{ f }+ K _{ f }$

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

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

$K =600 \,N / m$

Standard 11
Physics

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