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 5 kg and speed of 9 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Apr 15, 2016

$\approx 8.2 m$

Explanation:

By conservation of mechanical energy when the spring is compressed fully after collision
PE gained by the spring = Initial KE of the colliding object
$\implies \cancel{\frac{1}{2}} k {x}^{2} = \cancel{\frac{1}{2}} m {v}^{2}$
where
m = mass=5kg
v=velocity of the object=$9 \frac{m}{s}$
k = force constant=$6 \frac{k g}{s} ^ 2$
x= compression of spring=?
$\implies {x}^{2} = m {v}^{2} / k$
$\implies x = \sqrt{\frac{m}{k} {v}^{2}} = \sqrt{\frac{m}{k}} \times v = \sqrt{\frac{5 k g}{6 \frac{k g}{s} ^ 2}} \times 9 \frac{m}{s} \approx 8.2 m$