# Conservation of mechanical energy (no friction) - doesn't work in this case?

1. Jun 25, 2011

### Femme_physics

Conservation of mechanical energy (no friction) -- doesn't work in this case?

1. The problem statement, all variables and given/known data

http://img143.imageshack.us/img143/3181/springythingy.jpg [Broken]

In the drawing is described a lab device designed to eject balls upwards. The balls are ejected as a result from the spiral spring being compressed initially. In relaxed position to uppermost part of the spring is in line with the uppermost part of the device.

Given:

Ball weight = 0.5 [N]
Spring's hardness coeffecient = 3 N/cm
Initial compression of the spring = 12 cm

Calculate

A) The speed in which the ball is ejected from the device
B) The height H which the ball rises to

Ignore friction and air drag.

Comment: The cumulative energy in the elastic force of a spiral spring on runway h equals

http://img818.imageshack.us/img818/5626/echyj.jpg [Broken]

3. The attempt at a solution

I appear to have 2 unknowns in conservation of mechanical energy. I was told that when there's no friction I can use it to get the solution. What am I missing?

http://img818.imageshack.us/img818/8741/unnnnnnnnns.jpg [Broken]

Last edited by a moderator: May 5, 2017
2. Jun 25, 2011

### ehild

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

To make things clear, could you please list all variables (height of the ball and its velocity) and the energy terms
1. when the spring is compressed (initial state)
2 when the spring is relaxed
3 when the ball is at maximum height.

ehild

3. Jun 25, 2011

### I like Serena

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

Hi Fp! Glad to see you're still around!

Edited: as ehild said, you should break the problem down.

Last edited: Jun 25, 2011
4. Jun 25, 2011

### Femme_physics

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

Hi ehild, ILS!

Still around? When did I go anywhere?!?

What do you mean by "height of the ball"? I'm not given the radius

The compression is h1 -- The initial position is when the spring is at the height of the device. This says so in the question

That would be "H" in my formula, but again I have 2 unknowns in the conservation of mechanical energy

5. Jun 25, 2011

### ehild

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

Sorry, I meant the position of the ball. At what height is it.

Sorry, the question says that the spring has been compressed initially. Before that, it was not compressed, but then somebody came and pressed it, but that somebody and his/her force is out of question.
So initially the spring is compressed. Where is the ball, what its its velocity and what are the energy terms (kinetic, gravitational and elastic) ?

ehild

6. Jun 25, 2011

### I like Serena

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

Huh? :uhh:
I didn't say you went anywhere. I'm glad you're around!

Huh? I don't read that anywhere in the question?

7. Jun 25, 2011

### tiny-tim

Hi Femme_physics!

Each part should only involve one equation.

Can you write out the equations you're using, with just symbols instead of numbers, so it's easy to check?

8. Jun 25, 2011

### Femme_physics

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

I figured I can just define the height of the ball as my zero point. Can't I?

Well, it's potential energy in the form of elastic energy

Ah, not "still" around, but just "around" Always planning to be! :D

Sorry, I didn't mean the "initial" position but the "relaxed" position of the spring

9. Jun 25, 2011

### Femme_physics

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

I was using this:

http://img845.imageshack.us/img845/616/mgh.jpg [Broken][/QUOTE]

I just didn't have enough room to do it all in one line so I used two lines

Last edited by a moderator: May 5, 2017
10. Jun 25, 2011

### I like Serena

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

As I read it, the "relaxed" position is not the initial position.
Initially the spring is compressed.
Then it is released and uncompresses into its relaxed position.
After that the ball loses contact with the spring and jumps up.

Could you write the equation you just mentioned, applied to the initial compressed position and the relaxed position, just before the ball loses contact with the spring?

11. Jun 25, 2011

### tiny-tim

ok, then i can only see one unknown at a time …

if you know h2 = 0, the only unknown is v2

if you know v2 = 0, the only unknown is h2

12. Jun 25, 2011

### Femme_physics

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

Yes that's why I said sorry :)

I thought I did at my attempt at the solution. Should I make Vinitial to be 0? I was wondering that, because indeed H is my only unknown.

Is V1 also 0? same question as above

13. Jun 25, 2011

### I like Serena

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

Aha!!!
I think I just figured out what you did!

Yes, if you set V1 to 0, which it is, you'll only have 1 unknown left, which is H.

14. Jun 25, 2011

### ehild

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

Your life would be much easier with some systematic work. Let y the height where the ball is.

I. At the initial position of the ball: y=0, v=0, string compressed. Energy: (KE+PE+elastic)

II. relaxed state of the spring:
Ball at y = h1, v=v1, spring relaxed. Energy (KE + PE+elastic)

The ball lost contact with the spring.

III. At the highest poition y=H, v=0 no spring. Energy KE +PE.

The energy is the same at all positions, so you have one independent equation for v1 (EI=EII) and one for H (E1=EIII).

ehild

15. Jun 25, 2011

### Femme_physics

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

Ah, I'm an idiot! I pretty much wrote this solution before I posted the problem but I screwed up the math, but I thought this might be the case!!

Thanks for setting me straight!! :)

http://img40.imageshack.us/img40/8733/3333ru.jpg [Broken]

4.32 METERS.

Now Vo should be a piece a cake. I think.

http://img6.imageshack.us/img6/1966/vovovovo.jpg [Broken]

But I got the wrong Vo according to the answers. It should be:

9.08 m/sec

Last edited by a moderator: May 5, 2017
16. Jun 25, 2011

### I like Serena

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

Hmm, so you have a height H = 0.0432 m measured from the compressed position of the spring.
And the spring is compressed by h1 = 0.12 m.
Did you check all your units?

What did you use for delta y exactly?

17. Jun 25, 2011

### ehild

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

Check the unit of the stiffness constant.

ehild

18. Jun 25, 2011

### Femme_physics

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

I used 0.0432 meters for delta y

I always use meters, and since 12 cm is 0.12 meters, that's what I plugged in.

That'll be 3 N/m

(heh, you said "stiffness" *chuckles*)

19. Jun 26, 2011

### ehild

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

Sorry I meant the spring constant that you called "hardness coefficient". Read its value in the text of the problem, please. And you do it very well if you convert everything to SI units. Do with the spring constant, too.

ehild

20. Jun 26, 2011

### I like Serena

Re: Conservation of mechanical energy (no friction) -- doesn't work in this case?

What I meant was that the spring is compressed 12 cm, and when released it apparently comes up 4.32 cm.
That's not even out of the device!
At that point the spring is still compressed.
That can' t be right!

Errr, no?