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 `8 kg` and speed of `1 m/s` collides with and compresses the spring until it stops moving. How much will the spring compress?

Answer 1

Compression of the spring `=0.82m` rounded to second place of decimal.

Explanation:

Suppose the spring be compressed by `X` meter.

Force which compressed the string is the force applied by the moving body which came to rest after acting on the spring.
The speed of the object changed from `1ms^-1` to zero.
Using the expression from linear motion
`v^2-u^2=2as`,
where `v` is final velocity, `u` initial velocity, `a` acceleration and `s` distance moved.
Inserting the given values we obtain

`0^2-1^2=2aX`,

recall that the compressing objects moves distance `X` before it came to stop. This is the amount of compression of the spring produced by it. Solving for acceleration we obtain
`a=-1/(2X)`, `-ve` sign indicates that the acceleration is in a direction opposite to direction of movement of object or it is retardation.

We know that Force `F=ma`
`:. F=8xx(-1/(2X))=-4/X`

From Hooke's law for springs we know that
Reaction Force `F=-kX`, where `k` is the spring constant.
Now inserting the various values in the equation we obtain
`-4/X=-6X`, solving for `X`
`X^2=4/6`
or `X=sqrt(4/6)=0.82m` rounded to second place of decimal.