A spring with a constant of $3 \frac{k g}{s^{2}}$ $3 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $3 k g$ $3 kg$ 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 · 2017-09-06T05:41:10

The compression is $= 9 m$ $=9m$

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

The kinetic energy of the object is

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

The mass is $m = 3 k g$ $m=3kg$

The speed is $u = 9 m s^{- 1}$ $u=9ms^-1$

$K E = \frac{1}{2} \cdot 3 \cdot {(9)}^{2} = \frac{243}{2} J$ $KE=1/2*3*(9)^2=243/2J$

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

$P E = \frac{243}{2} J$ $PE=243/2J$

The spring constant is $= 3 k g s^{- 2}$ $=3kgs^-2$

So,

$\frac{1}{2} k x^{2} = \frac{243}{2}$ $1/2kx^2=243/2$

$x^{2} = \frac{243}{3} = 81 m^{2}$ $x^2=(243)/(3)=81m^2$

$x = \sqrt{81} = 9 m$ $x=sqrt(81)=9m$

A spring with a constant of 3kgs23 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 3kg3 kg and speed of 9ms9 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?