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 $4 kg$ and speed of $7 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-07-11T15:37:01

The spring will compress by $=5.3m$

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The spring constant is $k=7kgs^-2$

The kinetic energy of the object is

$KE=1/2m u^2$

The mass is $m=4kg$

The speed is $u=7ms^-1$

$KE=1/2*4*(7)^2=98J$

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

$PE=98J$

So,

$1/2kx^2=98$

$x^2=(2*98)/(7)=28m^2$

$x=sqrt(28)=5.3m$

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