Work Energy Power

JEE Main level, numerical answer, medium

89 of 183

Question 89 of 183

A 0.5kg0.5kg block moving at a speed of 12ms−112{ms}^{-1} compresses a spring through a distance 30cm30cm when its speed is halved. The spring constant of the spring will be _____ Nm−1{Nm}^{-1}

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Answer: 600

Given: Mass of the block m=500gm=500g, initial velocity of the block u=12ms−1u=12m{s}^{-1}, and final velocity of the block v=u2=6ms−1v=\dfrac{u}{2}=6m{s}^{-1}.
As non-conservative forces are not acting on the system we can conserve energy.
Thus, loss in kinetic energy of block == Gain in potential energy of spring
12mu2−12mv2=12kx2\dfrac{1}{2}m{u}^{2}-\dfrac{1}{2}m{v}^{2}=\dfrac{1}{2}k{x}^{2}
12×0.5×122−12×0.5×62=12k×0.32\dfrac{1}{2}\times 0.5\times {12}^{2}-\dfrac{1}{2}\times 0.5\times {6}^{2}=\dfrac{1}{2}k\times 0.{3}^{2}
k=600Nm−1k=600N{m}^{-1}

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