Gravitational Potential Energy of a skateboard

AI Thread Summary
The discussion focuses on ranking slopes based on the work done by gravitational force on a skateboarder during descent. The key equation used is the change in potential energy, represented as delta U = mg*(delta)h. It is clarified that since the change in height is the same for all slopes, the potential energy change is also identical across them. The initial assumption that the slopes should be ranked by their individual heights was incorrect. Ultimately, all slopes result in the same work done due to equal changes in height.
galtspeaking
Messages
3
Reaction score
0

Homework Statement



[PLAIN]https://dl.dropbox.com/u/9215647/webassign%20s.JPG

Rank the slopes (greatest first) according to the work done on the skater by the gravitational force during the descent on each slope,


Homework Equations



work=change in potential energy, ie. delta U = mg*(delta)h

The Attempt at a Solution



since potential energy is greatest at slope 1, shouldn't the rank be 1,2,3? but that is wrong..what am I doing wrong in this question then?
 
Last edited by a moderator:
Physics news on Phys.org
The change in height is the same for all three slopes. What does that tell you about the change in potential for each?
 
ah, I see..so they are all the same, b/c mgh is the same for all of them,

i was considering the total hight each of the slopes from the origin, i think that's where i was wrong
 
Correct!
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Is potential energy defined only for internal conservative forces?
Replies
15
Views
1K
Gravitational Potential Energy questions near the surface of a planet
Replies
8
Views
2K
Gravitational Potential Energy on an Incline
Replies
1
Views
2K
Is the line integral of tangential force in pendulum equal to the negative of change in potential energy?
Replies
3
Views
835
Misunderstanding of Gravitational Potential Energy
Replies
7
Views
3K
Gravitational potential energy traveling from earth to mars
Replies
7
Views
3K
Minimizing the Energy of Two Conductors Very Far Apart
Replies
23
Views
1K
Why Is Work Positive When Done Against the Electric Field?
Replies
22
Views
3K
Deriving gravitational potential energy -- mistake
Replies
9
Views
2K
Electrical Potential Energy and Potential
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top