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

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Answer 1 · 2017-06-01T05:51:17

The compression is $= 6 m$ $=6m$

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The spring constant is $k = 1 k g s^{- 2}$ $k=1kgs^-2$

The kinetic energy of the object is

$K E = \frac{1}{2} m u^{2}$ $KE=1/2m u^2$

$K E = \frac{1}{2} \cdot 4 \cdot {(3)}^{2} = 18 J$ $KE=1/2*4*(3)^2=18J$

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

$P E = 18 J$ $PE=18J$

So,

$\frac{1}{2} k x^{2} = 18$ $1/2kx^2=18$

$x^{2} = 2 \cdot \frac{18}{1} = 36 m^{2}$ $x^2=2*(18)/(1)=36m^2$

$x = \sqrt{36} = 6 m$ $x=sqrt(36)=6m$

A spring with a constant of 1kgs21 (kg)/(s^2) is lying on the ground with one end attached to a wall. An object with a mass of 4kg4 kg and speed of 3ms 3 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?