Question

A spring with a constant of `5 (kg)/s^2` is lying on the ground with one end attached to a wall. An object with a mass of `1 kg` and speed of `12 m/s` collides with and compresses the spring until it stops moving. How much will the spring compress?

Answer 1

The compression of the spring is about 5.4 m.

Explanation:

We can obtain the compression of the spring from the calculated energy balance between an instant just before the impact and the moment the object is stopped.

As the statement does not tell us otherwise, we must assume that there are no forces of friction. To solve the exercise with friction forces we would need more information (for example, the coefficient of dynamic friction of the object against the surface that supports the whole system).

In the absence of external forces, we know that mechanical energy is conserved. That is to say:

`Delta E = 0 color(white) {"...."} hArr color(white) {"...."} E (A) = E (B)`,

where `A` and `B` represent the moments before and after the shock.

At any instant of time, mechanical energy can be written as the sum of kinetic and potential energy.

Before the object collides against the quay, the object will have a kinetic energy due to its velocity and a gravitational potential energy that will depend on the height above the Earth's surface to which the experiment is performed.

For its part, given that the spring is initially at rest, its kinetic energy will be zero, the elastic potential will also be zero and will only have the gravitational potential energy that corresponds to it by the height to which it is, as it happens to the object.

We can develop the energy before the shock in the following way:

`E (A) = K (A) + U (A) = K (A) ("object") + U_{g} (A) ("object") +`#

`+ cancel {K (A) ("spring")} + U_{g} (A) ("spring") + cancel {U_{e} (A) ("spring")}`.

Once the object has hit the spring and it has been compressed, stopping it until it stops, the kinetic energy of the object becomes zero. Equally, the kinetic energy of the spring is also zero because at that moment it is at rest.

The gravitational potential energies of both the object and the spring will remain the same as those that had before the impact since the whole experiment is done on the ground, which we will consider horizontal.

`U_{g} (B) ("object") = U_{g} (A) ("object")`

`U_{g} (B) ("spring") = U_{g} (A) ("spring")`

Finally, the spring will have acquired a potential elastic energy as it is compressed by the object.

The calculation of the energy of the system after the collision and when the spring is completely compressed will be as follows:

`E (B) = K (B) + U (B) = cancel {K (B) ("object")} + U_{g} (B) ("object") +`

`+ cancel {K (B) ("spring")} + U_{g} (B) ("spring") + U_{e} (B) ("spring") =`

`= U_{g} (A) ("object") + U_{g} (A) ("spring") + U_{e} (B) ("spring")`.

Therefore, the mechanical energy balance will be as follows:

`E (A) = E (B) color(white) {"."} hArr`

`hArr color(white) {"."} K (A) ("object") + U_{g} (A) ("object") + U_{g} (A) ("spring") =`

`= U_{g} (A) ("object") + U_{g} (A) ("spring") + U_{e} (B) ("spring")`.

As we see, the gravitational potential energies cancel each other out by remaining constant throughout the experiment. Finally, we will have to:

`K (A) ("object") = U_{e} (B) ("spring")`.

Substituting the formulas of kinetic energy and elastic potential energy, we obtain:

`1/2 cdot m cdot v^2 = 1/2 cdot k cdot (Delta x)^2`.

From this expression can be cleared the compression of the spring:

`Delta x = v cdot sqrt {m/k}`,

and then we can calculate it:

`Delta x = 12 color(white) {"."} m/s cdot sqrt {{1 color(white) {"."} kg}/{5 color(white) {"."} {kg}/s^2}} ~~ 5.4 color(white) {"."} m`.