# A spring with a constant of 6 kgs^-2 is lying on the ground with one end attached to a wall. An object with a mass of 3 kg and speed of 9 ms^-1 collides with and compresses the spring until it stops moving. How much will the spring compress?

Mar 12, 2018

This is a case where kinetic energy ${E}_{k} = \frac{1}{2} m {v}^{2}$ is converted to spring potential energy ${E}_{s p} = \frac{1}{2} k {x}^{2}$.

${E}_{k} = \frac{1}{2} m {v}^{2} = \frac{1}{2} \times 3 \times {9}^{2} = 121.5$ $J$

Rearranging: $x = \sqrt{\frac{2 E}{k}} = \sqrt{\frac{2 \times 121.5}{6}} = 6.36$ $m$

#### Explanation:

Here is the rearranging of the spring potential energy equation. I have left off the subscript for simplicity:

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

Multiply both sides by 2:

$2 E = k {x}^{2}$

Divide both sides by $k$:

$\frac{2 e}{k} = {x}^{2}$

Swap the sides:

${x}^{2} = \frac{2 E}{k}$

Take the square root of both sides:

$x = \sqrt{\frac{2 E}{k}}$