Velocity and Distance involving springs

[eglaud]

Homework Statement

A spring with an elastic constant of 120 N/m is compressed 0.2 m, then launches a block of mass 0.5 kg on a table with friction 0.49, and it travels 0.9 m before stopping. If the block is then released at the edge of a table top 1.2 m above the ground, how far will the block travel?

Homework Equations

F = 1/2KX²
Vf² - Vi² = 2a(delta x)
Vf² - Vi² = -2g(delta y)

The Attempt at a Solution

I am stumped. I think these equations are relevant, but I'm not totally sure. I already solved for the friction coefficient, so some of those numbers may now be unimportant.

 

[kuruman]

If the block is then released at the edge of a table top 1.2 m above the ground, how far will the block travel?

This question is not very clear. Are we to assume that the block is at the end of the same spring that is compressed by the same amount before it is released?

 

[eglaud]

This question is not very clear. Are we to assume that the block is at the end of the same spring that is compressed by the same amount before it is released?

Yes sorry, the block is launched by the same spring off the table.

 

[haruspex]

Yes sorry, the block is launched by the same spring off the table.

Consider the launch velocity.

 

[eglaud]

Consider the launch velocity.

I don't know how to find the launch velocity. Do I set the force of the spring equal to kinetic energy?

 

[haruspex]

I don't know how to find the launch velocity. Do I set the force of the spring equal to kinetic energy?

Well, a force cannot equal energy, but that's nearly right.

 

[eglaud]

Well, a force cannot equal energy, but that's nearly right.

Well, I have absolutely no clue what I'm doingas finals week has left my mind in shambles. If it's nearly right, what can I do to tweak it and make it correct? I see that F = KE/deltaX, but in this case would the deltaX be the compression x or the x it travels?

 

[haruspex]

F = KE/deltaX

That is only valid for a constant force acting over distance x. When a spring expands or contracts a distance x the force is not constant.
Look to the relevant equations you quoted.

 

[eglaud]

That is only valid for a constant force acting over distance x. When a spring expands or contracts a distance x the force is not constant.
Look to the relevant equations you quoted.

Err, do I find it with the x equation? I have delta x but that's all. I'm afraid I'm not really grasping what you're hinting at.

 

[eglaud]

That is only valid for a constant force acting over distance x. When a spring expands or contracts a distance x the force is not constant.
Look to the relevant equations you quoted.

Would I set 1/2mv² equal to 1/2mk²? That's what I just found somewhere else, but I thought you told me that was incorrect (maybe my wording was just poor)

 

[haruspex]

Would I set 1/2mv2 equal to 1/2mk2?

Yes, set the PE in the spring equal to the KE produced.

thought you told me that was incorrect

No, you wrote

Do I set the force of the spring equal to kinetic energy?

That would mean an equation like kx=½mv².
As I replied, force is not energy.

 

[kuruman]

Would I set 1/2mv² equal to 1/2mk²?

Please note that 1/2mv² is the correct expression for the KE, but 1/2mk² is not the correct expression for the PE.

 

[haruspex]

Please note that 1/2mv² is the correct expression for the KE, but 1/2mk² is not the correct expression for the PE.

Thanks for catching that, I had not noticed.

 

[YarnMonkey]

Im not very sure what the question is asking although it seems that you are saying a spring compressed .2m with k=120 N/m launches a block off a table. The question, I think, is how far the block travels.

First we need to find the horizontal velocity of the block after the spring has relaxed and transformed all of its potential energy into kinetic energy.
We can write this as ½m(v*v) = ½k(Δx*Δx) where Δx = displacement of the compressed spring. (½k(Δx*Δx) = potential energy of the spring, PEs)
v = √k(Δx*Δx)/m
v = 3.1

Then, we need to find the time it takes for the block to fall. This can be done by the equation:
Δy = (vi*t)+½a(t*t)
In this case Δy is the change in height the object will experience (finale - initial) and we know it will land at a height of 0, so Δy = -1.2. We also know that a=-9.8 because we are finding the time it takes for the mass to fall. The initial velocity is 0 because it doesn't have any initial vertical velocity (velocity is a vector). Finally, this equation can be rearranged as:
t = √2*(-1.2)/-9.8
t = .495

This means that the mass will be in the air for a total of .495s before it hits the ground. During this time, if we neglect air resistance, the mass will have a constant horizontal velocity of 3.1m/s. This means that the distance the mass travels horizontal can be shown as:
Δx = v(t) +½a(t*t) where a=0 because there is no horizontal acceleration. This means that Δx = v(t) = 1.5m.

I hope that helped!

 

[CWatters]

Is there a diagram? Where is the edge of the table in relation to the relaxed length of the spring? Is some energy used up against friction before it's launched off the table?