# Elastic PE problem

1. Nov 25, 2011

### Nano-Passion

1. The problem statement, all variables and given/known data
I hate when I think I got the whole concept right only to get the answer wrong! I really hope it is a simple mis-calculation but I triple-checked all my work.

(a For the elevator of Example 7.9, what is the speed of the elevator after it has moved downward 1.00 m from point 1. (b)When the elevator is 1m below point 1, what is its acceleration? [I'm only concerned with part a for this problem, its the part that gives me trouble]

Here is all the information you need to know from example 7.9

m = 2000kg (19600 N)
initial velocity = 4 m/s
k = 10600 N/m
friction = 17000 N
Work of friction = -34000J

I set initial height to be 0m and height when compressed 1m to be -1m.
$$k_1+u_{g1}+u_{e1}+w_f=k_2+u_{g2}+u_{e2}$$
so then initial gravitational PE & initial elastic PE go to 0
$$1/2mv^2+w_f=1/2mv^2+mgh_2+1/2kx_2^2$$
$$16000J - 34000J = 1000 kg v^2 -19600J + 5300 J$$
since velocity is pointing down I put a - sign in front of it and I also add a couple terms.
$$1600J-34000 J= 1000kg (- v^2) - 14300 J$$
$$v=1.4 m/s^2$$

Edit: Corrected 10000kg to 1000 kg for 1/2m, but still getting a wrong answer.

Last edited: Nov 26, 2011
2. Nov 25, 2011

### Staff: Mentor

I think the issue is with the signs of the various energy components in your equation. It gets a bit complicated assigning them depending upon the side of the equation while taking the coordinate system directions into account. Let's take a look at where the energy starts and ends up.

Initially there's some kinetic energy in the descending elevator. It falls through 1m and thus gains energy mgΔh from gravity (Δh = 1m). Meanwhile it loses energy Ff*Δh to friction. In the end the remaining energy is in the KE of the elevator and PE of the spring which has just compressed by Δh=1m. Putting initial conditions with gains and losses on the left, and final conditions on the right,
$$\frac{1}{2}m v_o^2 + m g \Delta h - F_f \Delta h = \frac{1}{2}m v_1^2 + \frac{1}{2} k_s \Delta h^2$$
I think that sums it up. You can muck about with the signs of the terms if you want to make Δh negative or friction force negative, or move the terms from one side of the equation to the other.

3. Nov 26, 2011

### Nano-Passion

I don't think I did anything wrong with the negative and positive signs. Anyhow, I reran the calculation using your equation and I'm still getting the same answer. I've noticed that I entered 10000 as 1/2 mass at one point and corrected it; but all is in vain.

4. Nov 26, 2011

### Staff: Mentor

Strange, because I don't get your result.

5. Nov 26, 2011

### Nano-Passion

What result do you get? Correct answer according to cramster is 3.65 m/s.

I carefully plugged in all the numbers into your equations but its not the right answer.

6. Nov 26, 2011

### Staff: Mentor

My result matches Cramster's.

7. Nov 26, 2011

### Nano-Passion

Oh Gosh!!!!!

It is always the silly mistakes that catch me!!! I figured out what I did wrong this whole time since the beginning of the problem. Instead of assigning Work of friction to be 17000 I put in 34000. But the value of work of friction being 34000 was from the example that I was suppose to refer to.. of which was calculated for 2m!!! I needed the work done for 1m, but carelessly copied down the work from the example.

Very careless mistake.. Well thank you.

8. Nov 26, 2011

### Staff: Mentor

Glad it all worked out in the end

9. Nov 26, 2011

### Nano-Passion

Haha I'm very glad also. Every thread I've been creating here have been careless mistakes. Its funny how hard they are to spot sometimes haha. I really need to be careful of these mistakes, but I guess they won't happen in a test because I get too cautious; though they do suck up a lot of time at home!! I must have spent hours trying to see what is wrong when the problem should have taken me a few minutes.