# Help with Potential Energy vs Kinetic Energy and finding final velocity!

1. Dec 11, 2012

### riseofphoenix

Ok before we get to number 3, I need to show you what I did for number 2 because this is kind of a two-part problem:

PEinitial + KEinitial = [STRIKE]PEfinal[/STRIKE] + KEfinal
[STRIKE]m[/STRIKE]gy + (1/2)[STRIKE]m[/STRIKE]vi2 = (1/2)[STRIKE]m[/STRIKE]vf2
gy + (1/2)vi2 = (1/2)vf2
(9.81)(10) + (1/2)(102) = vf2/2
98.1 + 50 = vf2/2
148.1 = vf2/2
2(148.1) = vf2
296.2 = vf2
√(296.2) = vf
17.2 m/s = vf

So for this one, I didn't really know what to do :(
I tried doing this, but didn't get any of the choices above

PEinitial + KEinitial = [STRIKE]PEfinal[/STRIKE] + KEfinal
[STRIKE]m[/STRIKE]gy + (1/2)[STRIKE]m[/STRIKE]vi2 = (1/2)[STRIKE]m[/STRIKE]vf2
gx + (1/2)vi2 = (1/2)vf2
(9.81)(10) + (1/2)(252) = (1/2)vf2
98.1 + 312.5 = vf2/2
410.6 = vf2/2
2(410.6) = vf2
821.2 = vf2
√(821.2) = vf
28.6 m/s vf

But that's not right :(
Help?

2. Dec 11, 2012

### riseofphoenix

Oh wait!!!

1) If μ = 1.5, then fk = (1.5)(10 * -9.81)
fk = -147.15

2) If fk = -147.15, then F (forward) = +147.15

3) If F (forward) = +147.15, I can find acceleration, knowing that my mass, m, is 10 kg.

4) 147.15 = (10)a
147.15/10 = a
14.7 m/s2 = a

5) Then I tried plugging into a kinematic equation that had the 3 variables I had the values for (vf = 25 m/s, a = 14.7 m/s2, and Δx = 9.0 m)

vf2 = vi2 + 2axΔx
252 = vi2 + 2*14.7*9
625 = vi2 + 264.6)
625 - 264.6 = vi2
√(360.4) = vi
18.98 = vi
19.0 m/s = vi

Am I right? :D

Last edited: Dec 11, 2012
3. Dec 11, 2012

### SammyS

Staff Emeritus
Treat this as an entirely different problem in which the block is moving with a velocity of 25.0 m/s on the horizontal portion of the track, just before encountering the 9 meter long section having non-zero friction.

In your second post, you did calculate the friction force correctly.

Since you used energy concepts to solve problem #2, that seems like a good approach for this problem as well.
I suggest using the Work - Energy Theorem.​

By the way: The potential energy is constant in problem #3. --- Why ?

4. Dec 11, 2012

### riseofphoenix

oh well I said it was constant because of this diagram I found online.

5. Dec 11, 2012

### riseofphoenix

I have one more question though...

This one I have no idea how to solve for it... :(
It's a continuation of the last 2. It's basically the last part of this question.

Edit:

Oh wait!!!

Given
k = 50.4 N/m
vi = 10 m/s
m = 10 kg

KEinitial/moving block about to collide with spring in equilibrium = PEspring
(1/2)mv2 = (1/2)kx2
(1/2)(10)(102) = (1/2)(54)x2
500 = 27x2
500/27 = x2
18.5 = x2
√(18.5) = x
4.3 m = x

Is that right? :D

Last edited: Dec 11, 2012
6. Dec 11, 2012

### haruspex

Looks good.

7. Dec 11, 2012

### riseofphoenix

Thanks!

8. Dec 11, 2012

### SammyS

Staff Emeritus
That's sort of right .

Why did you call that vi ?

That's the speed after passing over the friction region.

9. Dec 11, 2012

### haruspex

Which is the starting point of part 3.