A spring with a constant of $12 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $6 kg$ 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 · 2017-04-08T06:17:48

The spring will compress by $=2.83m$

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The spring constant is $k=12kgs^-2$

The kinetic energy of the object is

$KE=1/2m u^2$

$KE=1/2*6*4^2=48J$

This kinetic energy will be stored in the spring as potential energy.

$PE=48J$

So,

$1/2kx^2=48$

$x^2=96/k=96/12=8$

$x=sqrt8=2.83m$

A spring with a constant of 12 (kg)/s^212 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 6 kg6 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?