• Support PF! Buy your school textbooks, materials and every day products Here!

Conservation of energy of an oscilating system with friction

  • Thread starter Oijl
  • Start date
  • #1
114
0

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!
 
Last edited:

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
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.
 

Related Threads on Conservation of energy of an oscilating system with friction

  • Last Post
Replies
9
Views
13K
Replies
30
Views
916
Replies
4
Views
2K
Replies
2
Views
4K
  • Last Post
Replies
3
Views
446
Replies
10
Views
4K
Replies
1
Views
3K
Replies
3
Views
6K
  • Last Post
Replies
2
Views
11K
Top