A spring with a constant of $12 \frac{k g}{s^{2}}$ $12 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $8 k g$ $8 kg$ and speed of $9 \frac{m}{s}$ $9 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

Question

1728 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-26T22:00:23

8 m

KE: Kinetic Energy = $\frac{1}{2} m v^{2}$ $1/2 m v^2$

PE: Potential Energy = $\frac{1}{2} k x^{2}$ $1/2 k x^2$

Energy Transformation:

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

$\frac{1}{2} (8 k g) {(9.81 \frac{m}{s^{2}})}^{2}$ $1/2 (8 kg) (9.81 (m)/s^2)^2$ = $\frac{1}{2} (12 \frac{k g}{s^{2}}) x^{2}$ $1/2 (12 (kg)/s^2) x^2$

Solving for x: compression,

x = 8 m

Note: in absence of gravitational force.

A spring with a constant of 12kgs212 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 8kg8 kg and speed of 9ms9 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?