Set 3 Problem number 17

A mass of 188 kg is raised 4.9 meters and at the same time stretches an ideal spring .93 meters from its equilibrium length. At this point the spring exerts a force of 418.5 Newtons.

Assume that on recoil the spring releases the object when the spring reaches its equilibrium position.

How much PE does the system gain during this process?
How does this example illustrate the nature of a conservative force?

Solution

Since the mass is raised through a positive distance, it does work against the opposing downward force exerted by gravity, which increases the PE of the mass. (The system consisting of the mass exerts its force in the direction of motion and therefore does work against gravity. Work done against a conservative force increases the PE of the system. )

The amount of work done against gravity is ( 188 kg * 9.8 m/s^2) * 4.9 meters = 1842.4 N * 4.9 meters = 9027.761 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 as was done against gravity in raising the object. In the process the elevation and therefore the PE of the object decreases and at the original position the additional PE has all been dissipated.

Thus in falling the increased gravitational PE is converted to kinetic energy.

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. If we walk across the room while lifting the object to shoulder level, then drop it to the floor, the result will be the same as if we lift it to shoulder level and drop it while standing in one place. Only the vertical lift matters.

In the process the mass stretches an ideal spring through a distance of .93 cm, exerting an average force of (0 + 418.5 Newtons) / 2 = 209.25 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 when stretched from the equilbrium position it exerts an average force equal to half its maximum).

The work done by the mass against the elastic restoring force is therefore

work against restoring force = Fave * `ds = 209.25 Newtons * 4.9 meters = 194.6025 Joules.

This work is the increase in the elastic PE of the system.
As the object returns freely 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, all of which contributes to the object's KE.

The total PE gain by the system is

Net PE gain = gain in gravitational PE + gain in elastic PE = 9027.761 J + 194.6025 J = 9222.363 J.

If the system is then released (assuming that at the instant the object returns to its original position the spring releases the object), as it returns to its original position it will gain as KE the 194.6025 J of elastic PE and the 9027.761 J of gravitational PE.

This will increase its KE by 9222.363 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 upon return to the initial position will be 9222.363 J, so that

.5 m v^2 = 9222.363 J and v = `sqrt(2 * 9222.363 J / m) = 1.438 m/s.

Note that in this case the forces exerted are all conservative:

The forces exerted against the gravitational and elastic forces as the mass is raised are equal and opposite to the forces exerted as the mass is lowered.
The net work done by the 'raising' force against the gravitational and elastic forces is therefore returned to the system in the form of KE.

Generalized Solution

In raising the mass a vertical distance `dy the work done against gravity was (m * g ) * `dy, where m is the mass of the object and g the acceleration of gravity. As the object falls the object encounters the identical gravitational forces it encountered in being lifted, and through the same distance. The work done by gravity on the object is therefore equal and opposite to the work done against gravity.

In stretching the spring through distance x, starting from equilibrium, we encounter average force (0 + kx) / 2 = kx / 2. This force is exerted through distance x so we do work F `ds = (kx / 2) * x = kx^2 / 2. As the spring returns to its equilibrium position the forces exerted on the object are equal and opposite to those exerted to stretch it, and the work kx^2 / 2 is done on the object by the spring.

The maximum total PE of the object is therefore

PEmax = m g `dy + kx^2 / 2,

as measured relative to its initial position.

Released from rest, when the object returns to its initial position its PE will have changed by - m g `dy - k x^2 / 2 so its KE will be m g `dy - k x^2 / 2 and we have

KE = .5 m v^2 = m g `dy + k x / 2.

This equation can easily be solved for the velocity v attained by the object.