# A spring with a constant of 7 (kg)/(s^2) is lying on the ground with one end attached to a wall. An object with a mass of 5 kg  and speed of  7 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Jul 11, 2017

The spring will compress by $= 5.9 m$

#### Explanation:

The spring constant is $k = 7 k g {s}^{-} 2$

The kinetic energy of the object is

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

The mass is $m = 5 k g$

The speed is $u = 7 m {s}^{-} 1$

$K E = \frac{1}{2} \cdot 5 \cdot {\left(7\right)}^{2} = 122.5 J$

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

$P E = 122.5 J$

So,

$\frac{1}{2} k {x}^{2} = 122.5$

${x}^{2} = \frac{2 \cdot 122.5}{7} = 35 {m}^{2}$

$x = \sqrt{35} = 5.9 m$