Question

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 `9 kg` and speed of `4 m/s` collides with and compresses the spring until it stops moving. How much will the spring compress?

Answer 1

Compression: `x = \sqrt{m/k}v_0 = \sqrt{(9\quad kg)/(6\quad N/m)}(4\quad m/s) = 4.899\quad m`

Explanation:

Mechanical Energy Conservation: When the forces acting on a system are purely conservative, its mechanical energy remains conserved.

`\Delta E = \Delta K + \Delta U = 0`

`\DeltaE` - Change in total mechanical energy;
`\DeltaK` - Change in kinetic energy;
`\DeltaU` - Change in potential energy;

In this case when the kinetic energy of the block decreases, the potential energy in the spring increases by the same amount to keep the total mechanical energy of the system a constant.

`m` - mass of the block; `\qquad v_0` - speed of the block.
`k` - spring constant; `\quad \qquad x` - compression of the spring;

Decrease in KE : `\Delta K = -1/2mv_0^2;`
Increase in PE : `\Delta U = + 1/2kx^2`

`\Delta E = 0 \qquad => \Delta U = - \Delta K`
`1/2 kx^2 = 1/2 mv_0^2; \qquad x = \sqrt{m/k}v_0 = \sqrt{(9\quad kg)/(6\quad N/m)}(4\quad m/s) = 4.899\quad m`