Set 3 Problem number 18

A mass of 49 kg slides up a frictionless incline until it is 4.3 meters higher than it was when it started. At some point the mass is 'caught' by an ideal spring, which it stretches 2.15 meters from its equilibrium length. At this position the spring exerts a force of 664.9 Newtons on the mass.

If the system is released from rest at this position, and if in returning its original position the system does 1695.57 J of work against some internal dissipative force or some external force that doesn't 'store' the energy, then

How much KE will the system gain by the time it returns to its equilibrium position?
How does this example illustrate the nature of a conservative force?

Assume that as the object returns to its original position the spring releases the object when the spring it at its equilibrium position.

Solution

Since the mass is raised through a positive distance, the system does work against the opposing downward force exerted by gravity.

The amount of work done against gravity is ( 49 kg * 9.8 m/s^2) * 4.3 meters = 480.2 N * 4.3 meters = 2064.86 J.
This is the increase in the gravitational potential energy of the mass.
As the mass falls back to its original position the gravitational force is in the direction of motion, so the same amount of work is done on the mass by gravity, so the gravitational PE is converted to kinetic energy and work against the dissipative force.

Note that only the vertical displacement of the object counts when determining the work done against gravity, since only the component of the displacement in the direction of the force has any effect on work or energy.

In the process the mass stretches an ideal spring through a distance of 2.15 cm, exerting an average force of (0 + 664.9 Newtons) / 2 = 332.45 Newtons against the elastic restoring force (note that for an ideal spring the force increases linearly from 0 at equilibrium to its maximum so that it exerts an average force equal to half its maximum).

The work done by the mass against the elastic restoring force is therefore Fave * `ds = 332.45 Newtons * 4.3 meters = 714.767 Joules.
This work is the increase in the elastic PE of the system.
As the object returns to its initial position the spring will exert an equal and opposite force, and an equal amount of work will be done on the object, part of this work contributing to the object's KE and part supplying the energy required to do the work against the retarding force.

The total PE gain by the system is therefore

net PE gain = 2064.86 J + 714.767 J = 2779.627 J.

If the system is then released, it will gain as KE the 2779.627 J (the 714.767 J of elastic PE and the 2064.86 J of gravitational PE), less the energy needed to do the work against the retarding force..

This will increase its KE by 2779.627 J - 1695.57 J = 1084.057 J as it returns to its original position (note that by the time the system has returned to its original position the spring will have returned to its equilibrium position).
If the system is released from rest its KE will be 1084.057 J, so that .5 m v^2 = 1084.057 J and v = `sqrt(2 * 1084.057 J / m) = 6.651 m/s.

The gravitational force and the spring force are conservative, in that then conserve energy--the work done to raise the object and to stretch the spring are 'stored' by virtue of vertical position and the stretch and tension of the spring. As the object returns to its initial position we reclaim this stored energy. The 1695.57 Joules which are 'lost' are are the result of work done against forces from which we cannot reclaim energy. These forces do not conserve energy and are called, appropriately, nonconservative forces.

If we denote the change in potential energy for the returning object `dPE, the change in its kinetic energy `dKE and the work done BY the system against the nonconservative forces as `dWnoncons, then we see that `dKE = -`dPE - `dWnoncons. This is equivalent to the general energy conservation equation

`dKE + `dPE + `dWnoncons = 0.