# A spring with a constant of 9 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 3 kg and speed of 5 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Mar 9, 2017

$\frac{5 \sqrt{3}}{3} m$

#### Explanation:

The object has kinetic energy, and assuming that the motion takes place horizontally, all this is given to the spring which stores it as Elastic Potential Energy

so

$K E = E P E$

$\frac{1}{2} m {v}^{2} = \frac{1}{2} k {x}^{2}$

$\cancel{\frac{1}{2}} \times 3 \times 5 \times 5 = \cancel{\frac{1}{2}} \times 9 \times {x}^{2}$

${x}^{2} = \frac{75}{9}$

$x = \sqrt{\frac{75}{9}} = \frac{\sqrt{75}}{3}$

$x = \frac{5 \sqrt{3}}{3} m$

of course we assuming that the natural length of the spring
$> \frac{5 \sqrt{3}}{3} m$