How Does Friction Affect the Energy of a Rolling Ball on Inclined Planes?

In summary: The initial energy is m⋅gh1, the final energy is KE + PE at the bottom of the table. You know the kinetic energy from the velocity you calculated above. The potential energy is m⋅gh2. The difference is the energy lost to friction. Use that and the work done by gravity to find the ratio.In summary, the ratio of work done by friction to work done by gravity can be calculated by finding the energy lost to friction, Wf, and the work done by gravity, Wg, which is equal to the change in potential energy. This can be done by solving for the horizontal and vertical components of the ball's motion and using the equations for kinetic and potential
  • #1
PhysicStuden7
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0

Homework Statement


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?


Homework Equations





The Attempt at a Solution


Wnc=(KEf-KEi)+(PEf-PEi)
 
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  • #2
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:
  • #3
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
Wf=mgcos(45)d
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.

Wg= mg[h1-h2]
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
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
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.
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.
 

1. What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed; it can only be transferred or converted from one form to another. This means that the total amount of energy in a closed system remains constant over time.

2. Why is conservation of energy important?

Conservation of energy is important because it helps us understand how energy behaves and how it can be utilized efficiently. It also allows us to predict and calculate the outcomes of various energy transformations, which is crucial for many scientific and technological advancements.

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Conservation of energy applies to everyday life in many ways. For example, when we turn on a light bulb, the electrical energy is converted into light energy; when we ride a bike, our muscular energy is converted into kinetic energy. Understanding this concept can also help us make more sustainable choices to conserve energy and reduce our carbon footprint.

4. Can energy be lost in a system?

No, energy cannot be lost in a closed system. It can only be transformed from one form to another. However, some energy may be lost to the surroundings as heat or sound, but it is not truly lost, as it is still present in the overall energy of the system.

5. Are there any exceptions to the law of conservation of energy?

There are no known exceptions to the law of conservation of energy. However, in certain situations, it may seem like energy is not conserved, but this is due to our limited understanding or measurement errors. The law has been extensively tested and has been proven to hold true in all cases.

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