# Energy/work on incline plane with springs

1. Nov 8, 2013

### Panphobia

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

Ok so the actual spring is attached to the 30 kg mass and is unstretched at this height off the ground. So if I were to pull the 25 kg mass down the plane 20 cm(assuming no friction on plane), and then let go, what would the velocity of the two masses be when the spring is unstretched again. The spring constant = 200(Light spring)

So here is my attempt(kind of), at a solution, I looked at the forces on both the 25 and 30 kg object, Let m1 = 25 kg and m2 = 30 kg
m1
ƩFx: T - m1gsin40 = m1a , normal force doesn't matter for this question

m2
ƩFy: m2g + kx - T = m2a

Ok now that I have done that I also did the energy on m2 since there is not enough information for energy on m1.

m2
Let h1 be 0.4m
Let h2 be 0.2m
Ei = Ef
(0.5)k*x^2 + m2*g*h1 = m2*g*h2 + (0.5)m2*v^2
100*x^2 + 58.8 = 15*v^2

After this I do not know what to do.

Last edited: Nov 8, 2013
2. Nov 8, 2013

### CAF123

In your final equation if mass m1 is pulled down 20cm, then mass m2 moves up 20cm assuming no slip of the string. This will tell you 'x' in your final equation.

Do you have the numerical answers to this problem?

3. Nov 8, 2013

### Panphobia

How does that tell me x? When x is the stretch occurring in the spring due to gravity?

4. Nov 8, 2013

### CAF123

Yes, x is the stretch of the spring. Since the spring is attached to the mass m2, if m2 moves up 20cm then the spring will stretch 20cm too.
I have answers for v1 and v2, but I would like to check them if possible - do you have the answer key?

5. Nov 8, 2013

### Panphobia

Oh yea, I had a brainfart, I thought the spring was not connected to the ground but only connected to the block. This was a question from a physics quiz we had, we didn't get the answers back yet. Also isn't v1 = v2 since they are connected? I am totally lost now because I did not incorporate the 25 kg block in this so it is as if the 30 kg is in free fall but not connected to another mass.

6. Nov 8, 2013

### CAF123

Your NII equation for m2 is missing a term. Knowing v2, you can find a2 using a kinematic equation. Then use the NII equations. I am not sure if v1 would necessarily equal v2, since m2 is always retarded by the spring force, at least when I calculate them, they are close, but not equal.

7. Nov 8, 2013

### Panphobia

since m2 is retarded by the spring force, wouldn't that make m1 the same, since they are connected in the same system. Also I updated the equation to ƩFy: m2g + kx - T = m2a.

8. Nov 9, 2013

### Tanya Sharma

Panphobia...are the two blocks at same height from the floor ? If it is then simply use energy conservation .That should give you the correct answer .

9. Nov 9, 2013

### CAF123

On second thought, they should have the same speed. If we denote the position of m2 by X and that of m1 by Y, then the constraint $X = -Y$ (I.e the no slip condition of the string) implies that $\dot{X}= -\dot{Y}$. I work them out to be close to each other, but not exact. What did you get for the speed v2?

10. Nov 9, 2013

### haruspex

Work conservation is certainly the simplest way, and I don't think it matters whether the blocks are the same height initially. You just have to assume the string does not become slack before the spring does, which seems reasonable.

11. Nov 9, 2013

### CAF123

Hi haruspex, do you agree that the speed of the masses should be the same for the reasons I gave in my previous post? We did actually use conservation of energy to find the speed of m2.

12. Nov 9, 2013

### Panphobia

Well in the actual question, it is never specified...The length of the incline isn't specified, and nothing about m1, except for its mass and that its pulled 20 cm. I am thinking of what I would have to do to get the velocity of that masses including the m1 in the system, but all I can think of won't work.

13. Nov 9, 2013

### Panphobia

If we are not given the initial height of the 25 kg block, and we are not given the height relative to the 30 kg block, how could I use energy for this completely, because energy would just give the velocity without the 25 kg block being connected to the 30 kg one.

14. Nov 9, 2013

### haruspex

The trouble with string is that it has no compressive strength. You have to be a bit careful sometimes and check that the string will not go slack at some point. I think it's reasonably obvious that it won't here, but I can't think of a trivial proof.
Yes, but unless I'm misreading it it was still a bit indirect.
The circumstances start and finish with the spring at its relaxed length. What does that tell us about the positions of the two masses in relation to their initial positions? So how about the change in PE?

15. Nov 9, 2013

### CAF123

My understanding is that if the mass m1 is pulled down by 20cm, then m2 will move up 20cm since we are assuming the string does not slip. So the motion starts with m2 at an initial height 40cm relative to the ground and it's final height is 20cm rel. to the ground when the spring is at its unstretched length. So ΔPE = m2g(20cm).

16. Nov 9, 2013

### haruspex

You're forgetting the change in PE of m1. The pulling down of m1 lowers the PE of m1, increases the PE of m2, and increases the PE of the spring. At the final position, all of those are back to initial state, so all that PE has gone into KE.

17. Nov 9, 2013

### Panphobia

Soooo the forces don't matter is what you are trying to say, at least in conservation of energy?

18. Nov 9, 2013

### haruspex

I wouldn't put it quite that way. I'm saying that conservation of energy is often a shortcut that allows you to skip analysing the forces and accelerations.

19. Nov 9, 2013

### Panphobia

So basically you do not need to know the height of m1? You just need to know how much energy was lost/gained? Also with this information, how would you get the total energy?

20. Nov 9, 2013

### Panphobia

I was thinking, and could you do the change in energy in m1 = the change in energy in m2, so the change in energy is just Einitial - Efinal....would this be correct or does that not work? I was thinking since they are connected.