Question

A spring with a constant of `4 (kg)/s^2` is lying on the ground with one end attached to a wall. An object with a mass of `2 kg` and speed of `3 m/s` collides with and compresses the spring until it stops moving. How much will the spring compress?

Answer 1

The spring will compress `1.5`m.

Explanation:

You can calculate this using Hooke's law:
`F=-kx`

`F` is the force exerted on the spring, `k` is the spring constant and `x` is the distance the spring compresses. You're trying to find `x`. You need to know `k` (you have this already), and `F`.

You can calculate `F` by using `F=ma`, where `m` is mass and `a` is acceleration. You're given the mass, but need to know the acceleration.

To find the acceleration (or deceleration, in this case) with the information you have, use this convenient rearrangement of the laws of motion:
`v^2=u^2+2as`
where `v` is the final velocity, `u` is the initial velocity, `a` is the acceleration and `s` is the distance travelled. `s` here is the same as `x` (the distance the spring compresses = the distance the object travels before stopping).

Substitute in the values you know
`v^2=u^2+2as`
`0^2=3^2+2ax` (final velocity is `0` as the object slows to a stop)
`a = \frac{-9}{2x}` (rearrange for `a`)
Notice that the acceleration is negative. This is because the object is slowing down (decelerating).

Substitute this equation for `a` into `F=ma`
`F=ma`
`F=m \frac{-9}{2x}`
`F=2\frac{-9}{2x}` (You know that `m=2`)
`F=\frac{-9}{x}` (The factor of `2` cancels)

Substitute this equation for `F` into the equation for Hooke's law:
`F=-kx`
`\frac{-9}{x}=-kx`
`x^2= \frac{-9}{-k}` (Rearrange for `x`)
`x^2 = \frac{9}{4}` (The minus signs cancel. You're given `k=4`)
`x= \frac{\sqrt{9}}{\sqrt{4}}` (Solve for `x`)
`x = \frac{3}{2} = 1.5`

As you're working in SI units, this distance has units of meters.
The spring will compress `1.5`m.