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 $3 \frac{m}{s}$ $3 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2016-05-13T13:04:21

$\sqrt{6} m$ $sqrt6m$

Consider the inital and final conditions of the two objects (namely, spring and mass):

Initially:
Spring is at lying at rest, potential energy = $0$ $0$
Mass is moving, kinetic energy = $\frac{1}{2} m v^{2}$ $1/2mv^2$
Finally:
Spring is compressed, potential energy = $\frac{1}{2} k x^{2}$ $1/2kx^2$
Mass is stopped, kinetic energy = 0

Using conservation of energy (if no energy is dissipated into the surroundings), we have:

$0 + \frac{1}{2} m v^{2} = \frac{1}{2} k x^{2} + 0$ $0+1/2mv^2 = 1/2kx^2+0$

$=>cancel(1/2)mv^2 = cancel(1/2)kx^2$

$=> x^2 = (m/k)v^2$

$:. x = sqrt(m/k)v = sqrt((8kg)/(12kgs^-2))xx3ms^-1 = sqrt(6)m$

A spring with a constant of 12 (kg)/s^212kgs212 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 8 kg8kg8 kg and speed of 3 m/s3ms3 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?