Conservation of energy of an oscilating system with friction

[Oijl]

Homework Statement

The cable of the 1800 kg elevator in Figure 8-56 snaps when the elevator is at rest at the first floor, where the cab bottom is a distance d = 3.7 m above a cushioning spring whose spring constant is k = 0.15 MN/m. A safety device clamps the elevator against guide rails so that a constant frictional force of 4.4 kN opposes the motion of the elevator.

Using conservation of energy, find the approximate total distance that the elevator will move before coming to rest.

Homework Equations

The Attempt at a Solution

In the beginning, the moment the cable snaps, the total energy of the elevator is its gravitational potential energy. If I want the total energy of the elevator to be only the elastic potential energy when the elevator has compressed the spring as much as it can the first time it falls onto the spring, then I would define the zero point for gravitational potential energy as that point of maximum compression.
With this setup, the total energy that the elevator has is mg(h+x), where h is the 3.7 meters above the equilibrium position and x is the maximum compression of the spring, known by calculation to be 0.901 meters.
All this gravitational potential energy is converted into elastic potential energy and heat energy due to friction. Then all that elastic potential energy is converted into gravitational potential energy and heat energy due to friction. Then all THAT gravitational potential energy is converted into elastic potential energy and heat energy due to friction... I want to find when the elevator is not moving, which is to say, when it has no more energy, except for the gravitational and elastic potential energies it would have from resting on the spring, which would be compressed some but not to the zero gravitational point. So, I figure I could just equate the total energy at the beginning to some gravitational and elastic potential energy and heat energy lost to friction over some distance d.
So I get
mg(h+x) = fd + mgc +(1/2)k(c^2)
where h is the initial height from the equilibrium position of the spring, x is the maximum compression of the spring, f is the force due to friction, d is the total distance traveled by the elevator, and c is the compression of the spring when the elevator is at rest.

SO, I think if I could find c, I could solve the problem. But how could I find c, considering that the elevator never actually is at rest, but continuously oscillates?

Sorry I'm so long-winded, but I wanted to describe my train of logic well. Thanks for reading this!

 

[Delphi51]

You may have a very clever way of solving this problem quickly, but it seems to me there is something too good to be true about it. That .901 is only good for the first fall. The elevator is going to bounce and then compress the spring less than .901 on the second fall. I think you will have to work out the first fall, then the bounce, then the second fall and see if there is any energy left for the 3rd bounce.

 

[nlightner]

I would first commend the student for their thorough thought process and attempt at solving the problem. However, I would also point out a few potential issues with their approach.

Firstly, while the student correctly identifies the initial total energy of the system, they do not take into account the work done by the frictional force during the first fall onto the spring. This work would reduce the initial total energy and should be accounted for in the equations.

Secondly, the student's equation for the total energy does not take into account the fact that the elevator will not only compress the spring, but also move up and down due to the oscillations. This means that the compression of the spring (c) will not remain constant and will vary with each oscillation.

Finally, the student's method of equating the total energy at the beginning with the potential and kinetic energies at the end is not applicable in this scenario, as the elevator does not come to a complete stop and there is no defined end point.

Instead, I would recommend approaching the problem using the principle of conservation of mechanical energy, which states that the total mechanical energy (potential and kinetic) of a system remains constant in the absence of external forces. In this case, the external force is the frictional force.

Using this principle, we can set up the following equation:

mg(h+x) - Wfriction = (1/2)k(c^2)

Where Wfriction is the work done by the frictional force, which can be calculated using the given force and the distance traveled.

Then, we can solve for c using this equation and plug it into the equation for the total distance traveled (d) given by:

d = 2(h+x) + 2c

This takes into account the fact that the elevator will travel up and down twice for each oscillation.

In conclusion, while the student's approach shows a good understanding of the concepts involved, it may not lead to an accurate solution in this particular scenario. Using the principle of conservation of mechanical energy and accounting for the work done by the frictional force should provide a more accurate result.