Question

A spring with a constant of `12 (kg)/s^2` is lying on the ground with one end attached to a wall. An object with a mass of `6 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 `2.12` metres.

Explanation:

We can either do this using conservation of energy or kinematics. I'll use conservation of energy.

The object's potential energy will be at `0` (because it's on the floor). It's kinetic energy will be given by `1/2mv^2`. Once it hits the spring, some of it's kinetic energy will be transferred to the spring. Therefore our equation will be

`1/2mv_"initial"^2 = 1/2mv_"final"^2 + 1/2kx^2`

We know that `v_"initial" = 3 m/s` and `v_"final = 0 m/s`. Therefore we get:

`1/2(6)(3)^2 = 1/2(6)(0)^2 + 1/2(12)x^2`

`27 = 6x^2`

`x = sqrt(27/6) ~~2.12 m`

Hopefully this helps!

Answer 2

`=>x=sqrt(4.5) " m" approx 2.1 " m"`

Explanation:

Quantities given:

• `k = 12 " kg" * "m/s"`
• `m = 6 " kg"`
• `v = 3 " m/s"`

We can calculate the kinetic energy of the object at the moment before it contacts the spring:

`=>E_k = 1/2mv^2`

The potential energy of a spring is given as:

`=>E_p = 1/2kx^2`

When the object comes to rest, it will be because the object has lost all of its kinetic energy and the energy has been transferred into the spring.

`=>E_k = E_p`

`=>1/2mv^2 = 1/2kx^2`

`=>mv^2=kx^2`

We can solve for `x`, which is the displacement of the spring (assuming its initial displacement was `0`):

`=>x = sqrt((mv^2)/(k))`

Substituting known values:

`=>x=sqrt(((6)(3)^2)/(12))=sqrt(4.5) approx 2.1 " m"`