A spring with a constant of $2/5 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $7/5 kg$ and speed of $5/4 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-05-02T21:02:14

The spring will compress by $=2.34m$

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The spring constant is $k=2/5kgs^-2$

The kinetic energy of the object is

$KE=1/2m u^2$

$KE=1/2*7/5*(5/4)^2=1.09375J$

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

$PE=1.03975J$

So,

$1/2kx^2=1.03975$

$x^2=1.03975*2/(2/5)=5.46875m^2$

$x=sqrt(5.46875)=2.34m$

A spring with a constant of 2/5 (kg)/s^22/5 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 7/5 kg7/5 kg and speed of 5/4 m/s5/4 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?