A 800-g block is dropped onto a relaxed vertical spring that has a spring consta

Question

A 800-g block is dropped onto a relaxed vertical spring that has a spring constant k =240.0 N/m. The block becomes attached to the spring and compresses the spring 52.5 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by the gravitational force on it?

While the spring is being compressed, what work is done on the block by the spring force?

What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)

If the speed at impact is doubled, what is the maximum compression of the spring?

Explanation / Answer

This is a question about energy.
All the kinetic energy of the block goes to the elastic potential energy of the spring. Note: its safer to use meters (SI unit) than cm when dealing with energy, but we can use cm here

KE=EPE
(1/2)mv^2=(1/2)kx^2
(1/2) x 0.8 x v^2= (1/2) x 240 x 52.5^2
v=909.3cm/s

Looking at the equation, if v is doubled, then compression must be doubled also (because both are squared). So doubling the velocity will give 105 cm max compression.

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