Conservation of energy of an elevator cable

In summary: For (b):I used 1/2k(x)^2 - F(x+h) = (1800)(9.81)(x+h)I plugged in x from (c), and got a number smaller than the original equation. Thanks!
  • #1
iamkristing
33
0

Homework Statement


he cable of the 1,800 kg elevator cab in Fig. 8-51 snaps when the cab 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 cab against guide rails so that a constant frictional force of 4.4 kN opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance x that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance (above the point of maximum compression) that the cab will bounce back up the shaft.


Homework Equations





The Attempt at a Solution



I got (a)...b and c are where I hit problems.

for b i used the equation...

mg(h+x) + F(h+x)(cos(pi))=.5kx^2

Now when I tried to solve for x i got an equation with huge numbers:

49012+13247x-0.075x^2
It just doesn't seem reasonable to solve for x here, and I know i must be doing something wrong I just can't see it.

for (c):

I used 1/2k(x)^2 - F(x+h) = (1800)(9.81)(x+h)

I would plug in x from c, its just x that I can't find.

Thanks!
 
Physics news on Phys.org
  • #2
iamkristing said:
for b i used the equation...

mg(h+x) + F(h+x)(cos(pi))=.5kx^2
Looks good.

Now when I tried to solve for x i got an equation with huge numbers:

49012+13247x-0.075x^2
Check those numbers. (Realize that MN = 1 000 000 N.)
 
  • #3
im still getting an answer of about 6 million...when the answer is 90...
 
  • #4
Check again.
iamkristing said:
Now when I tried to solve for x i got an equation with huge numbers:

49012+13247x-0.075x^2
The numbers in your first two terms are reasonable, but that 0.075 is WAY off. (It's a million times too small.)
 
  • #5
i adjusted that term by a million. and now that term is too large. I plugged in the correct answer of 90, hoping that the entire term will go to zero. But it goes to a number greater than -6 million.

I am supposed to be solving for the x^2 and the x in the equation correct?
 
  • #6
Not sure where that answer of 90 m comes from. I solved it, getting a much smaller number. (I might certainly have made an arithmetic error.)

Correct that last term and solve the quadratic equation.
 

1. How does the conservation of energy apply to an elevator cable?

The conservation of energy principle states that energy cannot be created or destroyed, only transferred from one form to another. In an elevator cable system, potential energy is converted into kinetic energy as the elevator moves up or down, and the kinetic energy is then converted back into potential energy when the elevator comes to a stop.

2. Does the conservation of energy affect the speed of the elevator?

Yes, the conservation of energy does affect the speed of the elevator. As the elevator moves up, potential energy is converted into kinetic energy, causing the elevator to accelerate. As the elevator moves down, kinetic energy is converted back into potential energy, causing the elevator to decelerate. The rate at which this conversion occurs determines the speed of the elevator.

3. How does the weight of the elevator affect the conservation of energy?

The weight of the elevator does not affect the conservation of energy principle. The amount of energy required to move the elevator is determined by the difference in potential energy between the starting point and the ending point. This energy is the same regardless of the weight of the elevator.

4. Can the conservation of energy be violated in an elevator cable system?

No, the conservation of energy cannot be violated in an elevator cable system. The law of conservation of energy is a fundamental principle of physics and is always upheld, even in complex systems like elevators. Any apparent violations of this law are due to external factors, such as friction or drag.

5. How does friction play a role in the conservation of energy for elevator cables?

Friction can play a role in the conservation of energy for elevator cables by converting some of the potential energy of the elevator into heat or sound energy. This can result in a slight decrease in the amount of potential energy converted into kinetic energy, and therefore a slight decrease in the overall efficiency of the elevator system. However, the conservation of energy principle still applies, as the total amount of energy involved in the system remains constant.

Similar threads

  • Introductory Physics Homework Help
2
Replies
47
Views
7K
  • Introductory Physics Homework Help
Replies
3
Views
309
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
408
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
910
  • Introductory Physics Homework Help
Replies
24
Views
892
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top