Solve Elastic PE Problem: Speed & Acceleration

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In summary: I was suppose to use. So now I'm stuck because my answer for 1m is 3.65 m/s which is different from what I got for 2000 kg when I used 10600 N/m.In summary, when the elevator is 1m below point 1, its acceleration is 16000J.
  • #1
Nano-Passion
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Homework Statement


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

Edit: Corrected 10000kg to 1000 kg for 1/2m, but still getting a wrong answer.
 
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  • #2
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,
[tex]\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 [/tex]
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
gneill said:
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,
[tex]\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 [/tex]
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.

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
Nano-Passion said:
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.

Strange, because I don't get your result.
 
  • #5
gneill said:
Strange, because I don't get your result.

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
Nano-Passion said:
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.

My result matches Cramster's.
 
  • #7
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
Glad it all worked out in the end :smile:
 
  • #9
gneill said:
Glad it all worked out in the end :smile:

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.

Thank you for your help.
 

1. What is Elastic Potential Energy (PE)?

Elastic Potential Energy (PE) is the energy stored in an elastic object, such as a spring or rubber band, when it is stretched or compressed. It is a type of potential energy that can be converted into kinetic energy when the object returns to its original shape.

2. How do you calculate Elastic PE?

The formula for calculating Elastic PE is: PE = 1/2 * k * x^2, where k is the spring constant and x is the displacement of the object from its equilibrium position.

3. What is the relationship between Elastic PE, speed, and acceleration?

Elastic PE is directly related to both speed and acceleration. As the elastic object is released and returns to its original shape, the potential energy is converted into kinetic energy, causing the object to gain speed. The acceleration of the object is also directly related to the spring constant, which determines the amount of force the object experiences.

4. How does changing the spring constant affect the Elastic PE problem?

The spring constant plays a crucial role in the Elastic PE problem. A higher spring constant means that the object will experience a greater force, resulting in a higher potential energy. Changing the spring constant can also affect the speed and acceleration of the object, as well as the amount of work required to stretch or compress the elastic object.

5. What are some real-life examples of Elastic PE?

Some common examples of Elastic PE include a bouncing ball, a stretched rubber band, a compressed spring, and a trampoline. These objects all store elastic potential energy when they are deformed and release it as kinetic energy when they return to their original shape.

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