# A spring with a constant of 2 (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?

Apr 19, 2016

$10 m$

#### Explanation:

Energy is conserved in this reaction.

Kinetic energy of the object is given by $\frac{1}{2} m {v}^{2}$, where $m$ is mass and $v$ is velocity, and potential energy stored in the spring is given by $\frac{1}{2} k {x}^{2}$, where $k$ is the spring constant and $x$ is the distance it compresses. If there is always the same amount of energy, then

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

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

Rearranging this to make $x$, the distance of compression, the subject (because the question asks for $x$),

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

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

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

Putting in values the question gives us, $k = 2 \frac{k g}{s} ^ 2$, $m = 8 k g$, $v = 5 \frac{m}{s}$, then

$x = \sqrt{\frac{8 \cdot {5}^{2}}{2}}$

$x = \sqrt{\frac{200}{2}}$

$x = 10 m$