How Can I Solve This Dynamics Problem Involving Friction and Inclined Planes?

  • Thread starter Thread starter JakePearson
  • Start date Start date
  • Tags Tags
    Dynamics
Click For Summary
The discussion centers on solving a dynamics problem involving an object moving up an inclined plane with friction. The user struggles with deriving the stopping distance equation, specifically d = u² / (2g(sinθ + μcosθ)). Participants suggest two approaches: using dynamics to analyze forces and acceleration, or applying conservation of energy to account for energy lost due to friction. The importance of showing work for better assistance is emphasized, as it allows others to provide targeted help. Overall, the thread highlights the challenges faced by students in understanding complex physics concepts and the collaborative effort to clarify these topics.
JakePearson
Messages
52
Reaction score
0
hey guys, I'm in the middle for revising for a dynamics paper in January. I'm in the middle of a 1st year BSc Physics degree. Looking through some past papers, there are questions that i can do, and there are questions that i am unable to do. These are usually question where you have to derive. I am going to write down one of the questions that are causing me problems and i was wondering if you could help me on how to answer it.

Q. An object moving along a horizontal surface reaches the foot of a plane inclined at an angle \vartheta to the horizontal with a speed u. If the coefficient of friction between the object and the plane is \mu show that the object will come to rest a distance d up the plane where
d = u2 / 2g (sin\vartheta) + \mucos\vartheta​


I am a diagnosed dyslexic, not using it as an excuse, i will always try my best, but this is not sinking in, can you please help.
 
Physics news on Phys.org
We have to calculate the net retardation.

If you have the mass of the body and it's initial velocity, you can calculate it's rate of retardation (which will be a function of the friction and gravity) and finally the time/distance at which v = 0.


Now...you got to manipulate the equation that comes to the proof.
 
i can appreciate you comment. But this is what i mean. This isn't sinking in. My lectures speak to me like this. Could you work through it with me. Cheers
 
Imagine I give you a similar problem but without an incline...just one dimensional linear motion. So in this problem you have an initial velocity u and a constant deceleration -a...derive an equation for me predicting the stopping distance in terms of just those two variables. Your problem uses the same basic derivation, but now you need to figure out what -a is in your problem.
 
JakePearson said:
Q. An object moving along a horizontal surface reaches the foot of a plane inclined at an angle \vartheta to the horizontal with a speed u. If the coefficient of friction between the object and the plane is \mu show that the object will come to rest a distance d up the plane where
d = u2 / 2g (sin\vartheta) + \mucos\vartheta​
Be careful with parentheses. The answer should be:
d = u2/2g(sinθ + μcosθ)

There are two ways to approach this problem:
(1) Using straight dynamics. Hint: What forces act on the object? What's is its acceleration as it goes up the ramp? Use kinematics.
(2) Using conservation of energy. Hint: How much energy is 'lost' due to friction?

Pick one of those approaches and give it a shot. Show your work and you'll get plenty of help.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
13
Views
2K
Replies
24
Views
3K
Replies
5
Views
2K
Replies
13
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
4K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
12
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K