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5.Work, Energy, Power and Collision
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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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