A spring with a constant of $6 \frac{k g}{s^{2}}$ $6 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $8 k g$ $8 kg$ and speed of $9 \frac{m}{s}$ $9 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-06-10T06:09:21

The compression is $= 6.93 m$ $=6.93m$

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The spring constant is $k = 6 k g s^{- 2}$ $k=6kgs^-2$

The kinetic energy of the object is

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

$K E = \frac{1}{2} \cdot 8 \cdot {(9)}^{2} = 144 J$ $KE=1/2*8*(9)^2=144J$

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

$P E = 144 J$ $PE=144J$

So,

$\frac{1}{2} k x^{2} = 144$ $1/2kx^2=144$

$x^{2} = 2 \cdot \frac{144}{6} = 48 m^{2}$ $x^2=2*(144)/(6)=48m^2$

$x = \sqrt{48} = 6.93 m$ $x=sqrt(48)=6.93m$

A spring with a constant of 6kgs26 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 8kg8 kg and speed of 9ms9 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?