# A spring with a constant of 6 (kg)/(s^2) is lying on the ground with one end attached to a wall. An object with a mass of 8 kg  and speed of  5 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Mar 5, 2017

$5.77 m$

#### Explanation:

Assuming no energy is lost, the energy stored by the spring will equal the kinetic energy beforehand.

Potential energy in a spring is

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

where $k$ is the spring constant and $x$ is the extension.

Kinetic energy is given by

$E = \frac{1}{2} m {v}^{2}$

where $m$ is mass and $v$ is velocity.

If all the energy is conserved between the two,

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

The question asks for the extension, $x$, so we rearrange to find $x$:

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

$k {x}^{2} = m {v}^{2}$

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

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

We know from the question that $k = 6 , m = 8$ and $v = 5$, so

$x = \sqrt{\frac{8 \times {5}^{2}}{6}} = \sqrt{33.3} = 5.77 m$