JEE MainPhysicsEasyWork Power & Energy

A block of mass 1 kg initially at rest is dropped from a height
h on the a spring of force constant k=50N/m. The maximum compression in
the spring is x=1 m then height hof the block is

Show answer and solution

Change in gravitational potential energy
= Elastic potential energy stored in-compressed spring

\Rightarrow mg(h + x) = \frac{1}{2}kx^{2}

\Rightarrow 1 \times 10(h + 1) = \frac{1}{2} \times 50 \times 1^{2}

\Rightarrow h = \frac{3}{2}

Source: Grass-G / P-06 Physics paper 14 October.docx

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