Physics help!

[addis4ever]

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Can someone please help me with this physics question, i tried it sooo many times that i think i exhausted all of my options. Thanx.
http://www.cramster.com/answers-jul...s-crates-top-springs-uncomp_609255.aspx?rec=0

 

[Pisiform]

F = kx = mg ... gives u 'm'

T = 2 pi sqrt m/k

 

[addis4ever]

thank u so much! i got to the point where i worked the problem so much that i started confusing myself. and the funny thing is ur approach was my first attempt!! 👍

 

[addis4ever]

but to go back to my my trials and errors on this problem, why couldn't conservation of energy be used in this problem? Because there are three types of energies, potential energy (spring), gravitational potential energy, kinetic energy (which is 0). the equation is 1/2*k*x2 = m*g*delta h. from there we can also solve for m. why is this approach wrong? thanx again

 

[Pisiform]

As you said 3 kind of Energies are involved:

Taking compressed position as a reference point: KE = 0, GPE = 0, EPE = max
Middle Position: KE = Max, EPE = 0, GPE = mgx

Not all of the EPE is converted to GPE but is divided among GPE and KE

so, EPE = GPE + KE

 

[MT Headed]

but to go back to my my trials and errors on this problem, why couldn't conservation of energy be used in this problem? Because there are three types of energies, potential energy (spring), gravitational potential energy, kinetic energy (which is 0). the equation is 1/2*k*x2 = m*g*delta h. from there we can also solve for m. why is this approach wrong? thanx again

F = kx = mg is not true (assuming the obvious x=L0-L). At the bottom of the bounce the force down is mg and the force up is kx, but they cannot be equal... the weight is going to bounce up, so its net force must be in the upward direction. kx > mg at the bottom of the bounce.

You are correct, this is an energy problem.

(1/2) k x^2 = mgh = mgx
(1/2) k x = mg ( <---- ah ha!!! F=kx=mg when x is half the distance of the bounce!!!)
m = (kx) / (2g)

T = 2 pi sqrt(m/k)
T = 2 pi sqrt(kx/2gk)
T = 2 pi sqrt(x/2g)

T is independent of spring constant in this scenario.
2, pi, and g are constants.
So as X grows, T grows.

Solution: sort them all by total distance moved (L0-L). Bigger distance = bigger time required.

 

[Pisiform]

F = kx = mg is not true (assuming the obvious x=L0-L). At the bottom of the bounce the force down is mg and the force up is kx, but they cannot be equal... the weight is going to bounce up, so its net force must be in the upward direction. kx > mg at the bottom of the bounce.

You are correct, this is an energy problem.

(1/2) k x^2 = mgh = mgx
(1/2) k x = mg ( <---- ah ha!!! F=kx=mg when x is half the distance of the bounce!!!)
m = (kx) / (2g)

T = 2 pi sqrt(m/k)
T = 2 pi sqrt(kx/2gk)
T = 2 pi sqrt(x/2g)

T is independent of spring constant in this scenario.
2, pi, and g are constants.
So as X grows, T grows.

Solution: sort them all by total distance moved (L0-L). Bigger distance = bigger time required.

I see what you did, but I don;t understand why you did not take KE into consideration. The way you setup the energies is that all of EPE is converted to GPE which is not true since we have KE also. We are dealing with 3 energies instead of two.

 

[MT Headed]

I see what you did, but I don;t understand why you did not take KE into consideration. The way you setup the energies is that all of EPE is converted to GPE which is not true since we have KE also. We are dealing with 3 energies instead of two.

There is no kinetic energy at the top (displacement=L0) because the weight is not moving. It's about to fall.
There is no kinetic energy at the bottom (displacement=L) because the weight is not moving. It's about to rise.

At spots in the middle there are varying amounts of kinetic energy, but this problem seems to focus on the two endpoints (x=L0 and x=L) so I didn't see any need to drag kinetic energy into the mess as well.

 

[Pisiform]

There is no kinetic energy at the top (displacement=L0) because the weight is not moving. It's about to fall.
There is no kinetic energy at the bottom (displacement=L) because the weight is not moving. It's about to rise.

At spots in the middle there are varying amounts of kinetic energy, but this problem seems to focus on the two endpoints (x=L0 and x=L) so I didn't see any need to drag kinetic energy into the mess as well.

I don't think that's true. We have three points Lo, +L, -L. And I also agree that the net force is in the upward direction at the compressed state, but that net upward force is the same as the net downward force (weight) Fnet = ma = kx . I did couple of similar questions before and they were not done this way. I will ask my prof. anyway to make sure.

 

[MT Headed]

I don't think that's true. We have three points Lo, +L, -L. And I also agree that the net force is in the upward direction at the compressed state, but that net upward force is the same as the net downward force (weight) Fnet = ma = kx . I did couple of similar questions before and they were not done this way. I will ask my prof. anyway to make sure.

Again, in the compressed state at the bottom of the bounce, if the Fup = Fdown, then the system would have no reason to bounce the weight up to its original position. The net force on the block would be zero, and the weight would not accelerate in any direction.

As the question is written, the unweighted spring is at length L0. A weight is placed on top of the spring while it is at rest at position L0, and then the weight is released. It bounces all the way down to L, and then bounces back all the way up to the original position L0. It will keep bouncing up and down in simple harmonic motion around a point halfway between L0 and L. It could not possibly bounce above position L0 into -L, because there is no source of energy to raise the block that high.

 

[Pisiform]

Again, in the compressed state at the bottom of the bounce, if the Fup = Fdown, then the system would have no reason to bounce the weight up to its original position. The net force on the block would be zero, and the weight would not accelerate in any direction.

As the question is written, the unweighted spring is at length L0. A weight is placed on top of the spring while it is at rest at position L0, and then the weight is released. It bounces all the way down to L, and then bounces back all the way up to the original position L0. It will keep bouncing up and down in simple harmonic motion around a point halfway between L0 and L. It could not possibly bounce above position L0 into -L, because there is no source of energy to raise the block that high.

So what I am getting is if the spring is Horizontal then: Equilibrium position = unstrained position (no mass attached)

but if the spring is vertical then: Equilibrium position =/= unstrained position, but between the compresses and unstrained.

is that right?

 

[MT Headed]

So what I am getting is if the spring is Horizontal then: Equilibrium position = unstrained position (no mass attached)

but if the spring is vertical then: Equilibrium position =/= unstrained position, but between the compresses and unstrained.

is that right?

You have the right idea. In the vertical system you could slowly lower the weight onto the spring, and only release it when the mg(down)=kx(up). The system would be perfectly still, in static equilibrium. The location where you would finally release the weight would be half way between L0 and L in the problem.

 

[Pisiform]

You have the right idea. In the vertical system you could slowly lower the weight onto the spring, and only release it when the mg(down)=kx(up). The system would be perfectly still, in static equilibrium. The location where you would finally release the weight would be half way between L0 and L in the problem.

Yes, I agree. I kept confusing horizontal spring system to vertical spring system. I thought I covered vertical spring system too but when I opened the prep books like TBR,NOVA etc .. they did not discuss it. Do you know if we need to know it for he mcat?

Thanks for the help btw, I found one more area which is weak -__-. Thank you

 

[cpfeiffer]

I've seen that problem before too. I think you just need to know about springs in general.
F = -kx and U = 1/2 kx^2. I think the vertical problem is an application. I saw something like that on a diag. I just took. So I would say "yes" but only as an application. It's like the spring on both sides problem I saw on a different diag.