A spring with a constant of $\frac{2}{3} \frac{k g}{s^{2}}$ $2/3 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $\frac{1}{5} k g$ $1/5 kg$ and speed of $\frac{1}{4} \frac{m}{s}$ $1/4 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-04-13T05:45:57

The spring will compress by $= 0.14 m$ $=0.14m$

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

The kinetic energy of the object is

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

$K E = \frac{1}{2} \cdot \frac{1}{5} \cdot {(\frac{1}{4})}^{2} = \frac{1}{160} J$ $KE=1/2*1/5*(1/4)^2=1/160J$

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

$P E = \frac{1}{160} J$ $PE=1/160J$

So,

$\frac{1}{2} k x^{2} = \frac{1}{160}$ $1/2kx^2=1/160$

$x^{2} = \frac{1}{80 k} = \frac{1}{\frac{2}{3} \cdot 80} = \frac{3}{160}$ $x^2=1/(80k)=1/(2/3*80)=3/160$

$x = \sqrt{\frac{3}{160}} = 0.14 m$ $x=sqrt(3/160)=0.14m$

A spring with a constant of 23kgs22/3 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 15kg1/5 kg and speed of 14ms1/4 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?