Energy/work on incline plane with springs

[Panphobia]

Homework Statement

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^2After this I do not know what to do.

 

[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?

 

[Panphobia]

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

 

[CAF123]

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

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?

 

[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.

 

[CAF123]

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.

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.

 

[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.

 

[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 .

 

[CAF123]

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.

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?

 

[haruspex]

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 .

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.

 

[CAF123]

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.

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.

 

[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.

 

[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.

 

[haruspex]

Hi haruspex, do you agree that the speed of the masses should be the same for the reasons I gave in my previous post?

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.

We did actually use conservation of energy to find the speed of m2.

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?

 

[CAF123]

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?

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).

 

[haruspex]

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).

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.

 

[Panphobia]

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

 

[haruspex]

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

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.

 

[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?

 

[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.

 

[andymars]

Personally, I would consider the two masses combined to be my system and find the amount of energy needed to pull them 20cm. I'd break this into two steps. First I'd find the energy needed to pull the blocks 20cm against the net force of gravity on the blocks. After I've found that, I would add in the energy that had to be stored in the spring while stretching it. This would be the total work needed to move the blocks 20cm. All of this energy will turn into kinetic energy. Use m=m1+m2 and v1=v2.

 

[haruspex]

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.

No, they're different masses and the heights they move are different.

 

[Panphobia]

Yea that's why it was a speculation, wasn't absolutely sure.

 

[Panphobia]

Like really I could solve this question with forces, but have almost no idea how to do it with energy, not much of a simplification/shortcut to me.

 

[Tanya Sharma]

Let us take reference level of P.E to be at ground i.e at ground h=0 .Let m1=25kg and m2=30kg .

m2 is at height 20 cm .Let m1 be at height 20+h . Let t=0 be moment when m1 is stretched by 20 cm down the incline .

1) What is the P.E of m1 at t=0?
2) What is the P.E of m2 at t=0?
3) What is the P.E of the spring at t=0?

 

[Panphobia]

This is when they have already been stretched right? Then Em1 = mgh and Em2 = 1/2kx^2 + mg(40)

 

[haruspex]

Let us take reference level of P.E to be at ground i.e at ground h=0 .Let m1=25kg and m2=30kg .

m2 is at height 20 cm .Let m1 be at height 20+h . Let t=0 be moment when m1 is stretched by 20 cm down the incline .

1) What is the P.E of m1 at t=0?
2) What is the P.E of m2 at t=0?
3) What is the P.E of the spring at t=0?

Sorry, Tanya, but I think that's little more complicated than necessary.
When m1 is pulled 20cm down the incline:
- what is the change in height of m1, hence its loss of PE?
- what is the change in height of m2, hence its gain in PE?
- what is the gain in the PE of the spring?
- how much PE has been gained in total?

 

[andymars]

I really think it's easier to figure out the work needed to pull the system of blocks against the net force from gravity on them: m1g-m2gsin(40). Since work is force times displacement, multiply that sum of forces by 0.2m to figure out the work necessary to move the blocks 20cm. You're also doing work to stretch the spring, though, and since the force the spring exerts is not constant, you will have to use E=1/2kx^2 to find the energy in the spring. Add the energy in the spring to (m1g-m2gsin(40))0.2m to find the total work done on the blocks. When released, they'll accelerate to some velocity given by 1/2mv^2 where m=m1+m2 and v is the same for both blocks. If you want to consider both blocks separately, though, you'll have to solve for the tension, and that will make it more complicated than it needs to be.

 

[Panphobia]

So confusing...I think I am supposed to solve it with only energy

 

[Tanya Sharma]

This is when they have already been stretched right? Then Em1 = mgh and Em2 = 1/2kx^2 + mg(40)

No...P.E of m1 is incorrect.

Pick a pen and a paper and try and determine the height of m1 at t=0 .Remember m1 is at height (20+h) initially .Now it moves 20 cm down the incline .This 20 cm down the incline causes a further decrease in height .

Draw a right triangle with hypotenuse 20 cm and angle 40° .What will be the length of perpendicular ? This is the decrease in the height .

What do you get ?

 

[andymars]

Yep. You should solve using energy. My last post described how I would do it. You can answer haruspex's questions to get the same result.

 

[Panphobia]

Yea I know I messed up I thought 20 was the height for a second. Messed up but it is mg20 sin40

 

[Tanya Sharma]

Yea I know I messed up I thought 20 was the height for a second. Messed up but it is mg20 sin40

Well...If the initial height of m1 is 20+h ,then height of m1 at t=0 will be (20+h-20sin40°) .

Do you agree ?

 

[Panphobia]

I do agree

 

[Tanya Sharma]

Good...

So the total mechanical energy of the system at t=0 =m₁g(20+h-20sin40°) + m₂g(40) + (1/2)k(20)²

Do you get the same expression ?