A spring with a constant of $7 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $2 kg$ and speed of $4 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-06-13T13:00:33

I got $214cm$

During this process the initial kinetic energy $K$ of the object will be converted into elastic potential energy $U_(el)$ ; so we have:

$K=U_(el)$

Or:

$cancel(1/2)mv^2=cancel(1/2)kx^2$

Where $x$ represents the compression of the spring. Rearranging:

$x=vsqrt(m/k)=4sqrt(2/7)=2.14m=214cm$

A spring with a constant of 7 (kg)/s^27 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 2 kg2 kg and speed of 4 m/s4 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?