How Is Energy Transferred in a Spring-Launched Block on an Inclined Plane?

[wanamaa]

Homework Statement

The spring in the figure shown has a spring constant of k. It is compressed a distance of x meters, then launches a block of mass m kilograms. The horizontal surface is frictionless, but the coefficient of kinetic friction for the block on the incline is µk. The vertical distance up the incline is h, and the incline makes an angle T with the horizontal. Give all answers in terms of k, x, m, µk, h, T, and g.

1. How much mechanical energy is in the system when the block is at rest with the spring compressed?

2. How much mechanical energy gets converted to thermal energy as the block travels the length of the incline?

3. How much kinetic energy does the block have when it reaches the top of the incline?

Here's the link to the picture: http://img21.imageshack.us/img21/6514/phys122.png


2. The attempt at a solution
for #1, I know that mechanical energy = Kinetic + Potential, so what I have down is ΔK + ΔU, which would be (K*ΔX) + mg, maybe? I'm most stuck on this one.

For number 2 I have thermal energy = Kmicro + Umicro, which according to the book is Ethermal = µk*mg*Δx*cos(T). This doesn't sound right to me though.

For number 3, I have that Kinetic energy = Potential + thermal, which I concluded, from the book, that kinetic = mg+(Fk * ΔX * cos(T)). Is this pretty close?

Any help would be awesome, I haven't done physics homework in almost a year, and strictly symbolic problems always throw me for a loop for some reason.

 

[tiny-tim]

Hi wanamaa!

The spring in the figure shown has a spring constant of k. It is compressed a distance of x meters, then launches a block of mass m kilograms. The horizontal surface is frictionless, but the coefficient of kinetic friction for the block on the incline is µk. The vertical distance up the incline is h, and the incline makes an angle T with the horizontal. Give all answers in terms of k, x, m, µk, h, T, and g.

1. How much mechanical energy is in the system when the block is at rest with the spring compressed?

Well, it's at rest, so that's just the spring energy.

2. How much mechanical energy gets converted to thermal energy as the block travels the length of the incline?

Energy lost to heat = work done by friction.

3. How much kinetic energy does the block have when it reaches the top of the incline?
…
For number 3, I have that Kinetic energy = Potential + thermal, which I concluded, from the book, that kinetic = mg+(Fk * ΔX * cos(T)). Is this pretty close?

You're less likely to make mistakes with plus and minus signs if you write it in full …

KE + PE before = KE + PE after + work done.

 

[wanamaa]

Thank you! That makes so much more sense now.

I didn't post the follow up question but here it is:

"Assume spring constant k = 1000 n/M, mass m = 0.200 kg, initial spring compression x = 0.15 m, coeffiction of friction µk = 0.20, vertical height of incline h = 2.0 m, and angle T = 45 degrees.

Consider the mechanical energy the block had at the bottom of the incline. At the top of the incline, what percent of that energy is: Potential energy, Thermal energy, and kinetic energy?

Also, what distance d does the block sail through the air?"

For the first part, I feel that there will be no potential energy, since the object is in motion and that the kinetic and thermal energies depend on the equation found above. Therefore, .20*cos(45 degrees)*0.15 = 0.0212. From there I'm kinda stuck, since this only gives me the work. for the second part, I don't really know how to even start that one. Again, it's been a while since I've used physics equations, any help again would be awesome and I'll try to figure it out on my own as well.

 

[tiny-tim]

Consider the mechanical energy the block had at the bottom of the incline. At the top of the incline, what percent of that energy is: Potential energy, Thermal energy, and kinetic energy?
…
I feel that there will be no potential energy, since the object is in motion

Sorry, that makes no sense …

PE depends on height, and so long as the height has increased, so has the PE

in fact that's the whole point of PE … you can find it just by knowing where you are, not how you got there or how fast you're going

(remember, KE + PE = constant … if we ignore work done … so as you go up, your KE must decrease, so your PE increases)

Also, what distance d does the block sail through the air?"
…
for the second part, I don't really know how to even start that one.

Find the KE, that gives you the speed, you know the angle (45º), so use the usual projectile equations to find the range.

 

[wanamaa]

oh ok, I guess I got potential energy and kinetic energy confused then, when I said there'd be no potential energy. What you said makes a lot of sense, actually. Thanks a bunch for your help! :)