# A spring with a constant of 9 kgs^-2 is lying on the ground with one end attached to a wall. An object with a mass of 2 kg and speed of 12 ms^-1 collides with and compresses the spring until it stops moving. How much will the spring compress?

Jun 11, 2017

The compression is $= 5.66 m$

#### Explanation:

The spring constant is $k = 9 k g {s}^{-} 2$

The kinetic energy of the object is

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

$K E = \frac{1}{2} \cdot 2 \cdot {\left(12\right)}^{2} = 144 J$

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

$P E = 144 J$

So,

$\frac{1}{2} k {x}^{2} = 144$

${x}^{2} = 2 \cdot \frac{144}{9} = 32 {m}^{2}$

$x = \sqrt{32} = 5.66 m$

Jun 11, 2017

The displacement of the spring, $x$, will be $4$ $m$.

#### Explanation:

The kinetic energy of the moving object will be ${E}_{k} = \frac{1}{2} m {v}^{2} = \frac{1}{2} \times 2 \times {12}^{2} = 144$ $J$.

This will be converted to spring potential energy: ${E}_{\text{sp}} = \frac{1}{2} k {x}^{2}$.

Rearranging: $x = \sqrt{\frac{{E}_{\text{sp}}}{k}}$

We can substitute in the calculated value of ${E}_{k}$ for ${E}_{\text{sp}}$:

$x = \sqrt{\frac{{E}_{\text{sp}}}{k}} = \sqrt{\frac{{E}_{k}}{k}} = \sqrt{\frac{144}{9}} = \frac{12}{3} = 4$ $m$