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

Apr 13, 2017

#### Answer:

The spring will compress by $= 0.14 m$

#### Explanation:

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

The kinetic energy of the object is

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

$K E = \frac{1}{2} \cdot \frac{1}{5} \cdot {\left(\frac{1}{4}\right)}^{2} = \frac{1}{160} J$

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

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

So,

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

${x}^{2} = \frac{1}{80 k} = \frac{1}{\frac{2}{3} \cdot 80} = \frac{3}{160}$

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