An 80.0-kg man jumps from a height of 2.50 m onto a platform mounted on springs.

Question

An 80.0-kg man jumps from a height of 2.50 m onto a platform mounted on springs. As the springs compress, he pushes the platform down a maximum distance of 0.240 m below its initial position, and then it rebounds. The platform and springs have negligible mass.

* What is the man's speed at the instant he depresses the platform 0.120 m?

* If the man just steps gently onto the platform, what maximum distance would he push it down?

Explanation / Answer

a) Conservation of energy: m.g.h = (1/2) k.x^2 => k = 2mgh/x^2 Conservation of energy again: mgh' = (1/2)mv^2 + (1/2)k.x'^2 => v b) Step gently onto the platform, maximum distance is when he is in balance: mg = k.x => x = mg / k

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