# A spring with a constant of 6/7 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 3/5 kg and speed of 5/3 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Feb 27, 2016

$x = 1.3944 m$

#### Explanation:

This question used the concept of conservation of energy.
So that means, the total kinetic energy of the object has to be converted completely to spring energy since it stops.
i.e $\frac{1}{2} m {v}^{2} = \frac{1}{2} k {x}^{2}$

So, to find the total compression of the string, re-arrange the given formula into $x = v \setminus \sqrt{\frac{m}{k}}$

We have $v = \frac{5}{3} m {s}^{-} 1$, $m = \frac{3}{5} k g$, $k = \frac{6}{7} k g {s}^{-} 2$

So, $x = \frac{5}{3} \setminus \sqrt{\frac{\frac{\cancel{3}}{5}}{{\cancel{6}}^{2} / 7}} = \frac{5}{3} \cdot \setminus \sqrt{\frac{7}{10}}$

Solving using a calculator (never us it all the time), we get the above given answer justified to being so weird.