A spring with a constant of $1 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $2/3 kg$ and speed of $1/4 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-08-31T06:45:59

THe compression is $=0.289m$

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The spring constant is $k=1kgs^-2$

The kinetic energy of the object is

$KE=1/2m u^2$

The mass is $m=4kg$

The speed is $u=1/4ms^-1$

$KE=1/2*2/3*(1/4)^2=1/48J$

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

$PE=8J$

The spring constant is $=1kgs^-2$

So,

$1/2kx^2=1/48$

$x^2=(2)/(28)=0.0833m^2$

$x=sqrt(0.0833)=0.289m$

A spring with a constant of 1 (kg)/s^21 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 2/3 kg2/3 kg and speed of 1/4 m/s1/4 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?