A 2 kg block is attached to a horizontal ideal spring with a spring constant of 200N/m.when the spring has it`s equilibrium length the block is given a speed of 5m/s.what is the maximum elongation of the spring ?

Question

Vasanth SR · Answer

At the point where it has a speed of 5 m/s there is only kinetic energy and at max elongation there is only potential energy. So I set 1/2*2*5^2 = 1/2*200*x^2 and I solved for x and I got 0.5 m.

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A 2 kg block is attached to a horizontal ideal spring with a spring constant of 200N/m.when the spring has it`s equilibrium length the block is given a speed of 5m/s.what is the maximum elongation of the spring ?

A 2 kg block is attached to a horizontal ideal spring with a spring constant of 200N/m.when the spring has it`s equilibrium length the block is given a speed of 5m/s.what is the maximum elongation of the spring ?

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