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Edexcel M3, EPE on springs/strings etc watch

1. Can someone do the following problem, since I cannot do anything do with this. When answering, can you go through it stage by stage please, explaining what you are doing at each stage and why. Bear in mind I don't understand energy, and so this "Work-energy principle" i read about is news to me, please explain!

A typical question for you to solve, that I can't, is:

1) A jack-in-the-box is made using a spring of natural length 0.2m and modulus 100N and a 'jack' of mass 0.5Kg. When the lid is closed, the spring is compressed to a length of 0.1m. Assuming the spring to be vertical throughout, calculate the maximum distance that the 'jack' will rise when the lid is suddenly raised.

2) A particle P of mass 0.3Kg is suspended by two identical elastic strings of natural length 1.5m and modulus lambda N. The other ends of the strings are fixed to two points A and B on a horizontal ceiling where AB=3m. P is released from rest at the mid-point of AB and falls vertically until it is instantaneously at rest at a point 1m below the level of the ceiling. Calculate the value of lambda N and the speed of P when it is 0.5m below the level of the ceiling.

Any help would be appreciated

Kirk
2. 1 can be done be conservation of energy. This principle states that all energy an object has is conserved and that only changes in the type of energy it has.

It is a good idea to write down the intial energy the object has in terms of its KE, PE, and EPE. Then take a general position with extension x, write don the EPE and PE. In 1 when it is a maximum height it has KE = 0.
3. 1) the potential energy stored in a spring is λx²/2l (this is a formula you should know). so when you comprress the spring 0.1 m, you give it (100*0.1²/2(0.2)) = 2.5J of energy. some of this energy is transferred to GPE when the jack rises. it rises (for simplicity) a height 0.1 + h (the 0.1 is there because it was compressed to 0.1). at the top, the spring will have energy 100*h²/2*0.2 and the jack will have GPE mgh = 0.5*9.8*(0.1+h), and this will all equal 2.5J due to the conservation of energy

thus:

250h² + 4.9h -2.01 = 0

you solve this to get h=0.08 m (or h=-0.1 which corresponds tot he jack not moving at all)
4. But what is PE? Is it mgh? Or is it mg multiplied by the change in height?

I.e. if something is 5m off the floor, does it have PE of 5mg? If I move it up to 7m above the floor, is the new PE 7mg? ---- OR is it change in height, so change in height being 2m, so PE is now 2mg?

I seem to remember deltas in physics...but not in maths. I'm confused. What you said was a little help - perhaps could you elaborate more and do the questions so i can learn from the kinds of steps to take?

Kirk
5. (Original post by hatton02)
I.e. if something is 5m off the floor, does it have PE of 5mg? If I move it up to 7m above the floor, is the new PE 7mg?
Yes. but it depends on you defining that the floor has a PE of 0

if you move up from 5 to 7 m, the change in PE is 2mg.
6. The answer to question 1 is: h=0.180m

That is not the answer that was previously displayed...did you make an error? Your solution seemed far too complicated, h+0.1 - that came out of nowhere?

Can anyone make clear step by step solutions PLEASE!

Kirk
7. Draw a diagram with general position extension h

Initially the spring has the following energy:

KE = 0
PE = 0
EPE = 100*0.1*0.1/2*0.2 = 2.5J

So initial total energy = 2.5J

At its max height, the extension = h

KE = 0
PE = (0.1 + h)*0.5*9.8
EPE = 100h^2/2*0.2 = 250h^2

Initial energy = final energy. To get the equation elpaw did. This gives h = 0.08, but the max height is 0.1 + h = 0.18
8. (Original post by hatton02)
The answer to question 1 is: h=0.180m

That is not the answer that was previously displayed...did you make an error? Your solution seemed far too complicated, h+0.1 - that came out of nowhere?

Can anyone make clear step by step solutions PLEASE!

Kirk
sorry, yeah, i thought h was the height above the equilibrium position, but i guess the question wanted just the height risen. 0.18 = 0.1 + 0.08, i.e. the whole 0.1+h part.
9. I'm confused why 0.1+h is used? I'm also confused why initial PE = 0.
Surely it is 0.1m off the floor and so has some PE, i.e. mxgxh= 0.5x9.8x0.1 J?

Anyway, I calculated it as follows, can anyone point out my mistake? (This is why i hate mechanics so much, since it makes no sense and i don't know when i've gone wrong!) Grrr.

l=natural length=0.2m
lambda = 100N
m = 0.5Kg
x = extension/compression = 0.1m

EPE = lambda x²/2l = 2.5J

Gain in PE = Loss in EPE

At the maximum height, the amount of EPE will be 0. Therefore, since it has 2.5J of EPE now, the loss of EPE between start of motion and end of motion is 2.5J.

So: mgh = 2.5
0.5x9.8xh = 2.5

therefore h=0.51m

Which is wrong! Help!
Kirk
10. being 0.1m off the floor doesnt matter. it all depends on where you define the 0 of PE to be. i could for instance define it to be on the moon, in which case it will have -141376442mg PE. all that matters is the change in PE. to make calculating the change in PE easier, i defined 0 where it was in the 0.1m compressed stage.
11. (Original post by hatton02)
At the maximum height, the amount of EPE will be 0.
This bit is wrong. after it goes through the equilibrium position, it will gain EPE due to stretching.

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Updated: May 10, 2004
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