# Energy Equation Problem

1. Mar 26, 2009

### Ut Prosim

1. The problem statement, all variables and given/known data

The 10kg block has an initial velocity of 2m/s at the bottom of the ramp and slides up the ramp with a constant applied horizontal force F=75N as shown. Find the maximum height h that the block can reach on the ramp.

We're using cos(theta/%)=.8 & sin(theta/%)=.6

m=10kg | V1=2m/s | F=75N | Y2=? | Y1=0 | coefficient of friction(#)= .4

2. Relevant equations

K1 + Ug1+ Ue1+ W'1-2= K2 + Ug2 + Ue2

3. The attempt at a solution

K1= .5m(V1)2 V1=2 ==20
K2= .5m(V2)2 V2=0 ==0
Ug1= mgY1 Y1=0 ==0
Ug2= mgY2 = 98Y2

(elastics are 0)

W'1-2
= WF+Wf
= [-#*mgcos(%)*(y2/sin(%))] + [Fcos(%)*(Y2/sin(%))]
=[(-.4)(10)(9.8)(.8)*(Y2/.6)] + [(75)(.8)*(Y2/.6)]
=47.733Y2

Energy Equation:

20+0+0+47.733Y2 = 0 + 98Y2 + 0
Y2=.398

If you could review my work, and help me find out where I went wrong, it would be greatly appreciated.

2. Mar 26, 2009

### KLoux

Well... I don't get your answer, and I also don't get 0.249 m. The difference between my answer and yours comes from the energy equation and they way work is calculated. It looks like in your equation, the work done by friction helps and the work done by the external force hurts...

But like I said, I still don't get 0.249 m. It's always possible that I'm doing something wrong, too.

Also, it helps to use LaTeX for readability (it's hard to look at % and #... especially when you know it could be $$\theta$$ and $$\mu$$). I might be nitpicking, but I also like to use the same variables that you're given in the problem. If it asks you to solve for h, why change it to $$Y_{2}$$?

-Kerry

3. Mar 26, 2009

### Ut Prosim

The work done by friction hurts, hence it has negative mu . I just have them switched in the addition problem above.

I have no clue what LaTeX is, although I see when I try to copy and paste your mu, LaTeX Code: \\mu, it shows up as an odd code. I'm new to these forums and have not been introduced to this character code.

4. Mar 26, 2009

### LowlyPion

I don't think you have enough terms. Also I think solving for the distance up the ramp is the way to go and then figure the h after you know how far up it goes.

To do this figure the net force on the block on the ramp:

Total-F = Applied-F*Cosθ - mg*sinθ - Friction-F

First determine what's happening with the first 2 forces.

75*.8 - 98*.6 = 1.2 N - friction

Now Friction is

μ*(75*sinθ + 98*cosθ) = .4(75*.6 + 98*.8) = 49.36

Net Force then = 1.2 - 49.36 = -48.16

From your force now you have the deceleration and can figure the distance by

V2 = 2*a*x

Then just figure h from x.

5. Mar 27, 2009

### KLoux

For LaTeX help, click on the $$\Sigma$$ to the right of the array of buttons when you're postion.

LowlyPion found our mistake (I made the same one you did...): We didn't include the effect of the applied force F on the normal force when we calculated the friction force. If you do that, you should get the right answer (the same one you'd get with LowlyPion's method).

-Kerry