# Homework Help: Conservation of Energy

1. Oct 20, 2011

### PhysicStuden7

1. The problem statement, all variables and given/known data
A ball with mass, m is released from rest at a height, h1 and is allowed to roll down a ramp. After wrapping around a loop, it rolls up an incline of angle theta to height, h2, from which it launches. We measure the horizontal distance, d, it travels. To investigate the relative effect of friction, find the ratio of the work done by friction, Wf, to the work done by gravity, Wg.

The parameter is set to
m=85 [g], h1=1.00 [m], h2=0.53 [m], theta=45 degrees

The measurement is d= 30 [cm]

The ratio of work done by friction to work done by gravity is
Wf/Wg= ? (This is what I'm looking for)

Is this larger or smaller than you expected?

2. Relevant equations

3. The attempt at a solution
Wnc=(KEf-KEi)+(PEf-PEi)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 20, 2011

### Delphi51

Welcome to PF!
Wow, a complex problem for your first post.
It isn't entirely clear what Wg is. I would guess it means simply mg*h1, the total energy delivered by falling from the starting point to the ending point where it hits the table at height zero. That's easy to calculate.

The work against friction is what is left after you do mg*h1 - ending KE. So the core of your solution must be to find the horizontal and vertical speeds when the ball hits the table.
(have you done rotational motion yet? If so, you should include rotational kinetic energy. If not, forget it.)

I would write up the 2D motion problem in the usual way, assuming the initial speed is v. Separate that into horizontal and vertical components. Write your d=vt for the horizontal motion and a d= and a v= equation for the vertical motion. Put in all the knowns in all three equations and find the one where there is only one unknown so you can solve for it. If there isn't one, solve the d=vt equation for v and substitute into the others to get an equation with one unknown. Once you have t, you can find the velocities.

Good luck!

Last edited: Oct 20, 2011
3. Oct 20, 2011

### PhysicStuden7

So far this is my solution, and I have no idea if it is right. If anyone can help, please let me know.

Wf=mgcos(45)d
Wf=(0.085kg)(9.8m/s)(cos45)(0.3m)
Wf=0.177

Work by gravity= change in PE
Wg= mg[h1-h2]
Wg= (0.085kg)(9.8m/s)(1m-0.53m]
Wg=0.392

Wf/Wg= 0.177/0.392 = 0.452

4. Oct 20, 2011

### Delphi51

This doesn't make sense to me. There is no friction when traveling the distance d through the air. The friction occurs on the two ramps and the loop.

This doesn't look right because the ball falls from h1 to the ground; its flight doesn't end when it is at the top of the second ramp at height h2.

5. Oct 21, 2011

### PhysicStuden7

I figured that was probably wrong, but I'm not really sure where to go from here. Can you point me in the right direction of where to begin? I understand that the work done by gravity is m*g*cos(theta), but I am confused on how to figure out the work done by friction. Any help would be appreciated.

6. Oct 21, 2011

### Delphi51

Let v = speed at top of second ramp. Vx=v*cos(45)=.707v = Vy
Horizontal: x = vt -> 0.3 = .707vt [1]
Vertical: y = Vi*t + ½a⋅t² -> -.53 = .707vt - 4.91t² [2]
Sub [1] into [2] and solve for t, the time of flight from the ramp to ground. Sub back into [1] to get v.
Use the vertical velocity equation to find the final vertical velocity when it hits the ground. Combine Vx and Vy to get the total velocity when it hits the ground. Then calculate the ending kinetic energy.
The initial energy minus the final energy is the energy lost to friction, Wf.