Non conservative forces: work + coefficient of static friction

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[SOLVED] Non conservative forces: work + coefficient of static friction

hey guys, I've been having a little trouble with some parts of this problem, hope someone can point me in the right direction



A 20[kg] sled starts up a 30 degrees incline with a speed of 2.4[m/s] the coefficient of kinetic friction is 0.25.

A) How far up the incline does the sled travel?

B) What condition must you put on the static coefficient if the sled is not to get suck at the point determined in part a?

C) If the sled slides back down what is its speed when it returns to its starting point?

Know: m = 20 [kg]
theta = 30 degrees
v (initial) = 2.4 [m/s]
Mu(k) = 0.25




Ok, so I did part A): I found the net work and how far the sled went up the incline.

Part B) threw me off a little bit. By condition I assume I have to find an angle? Or maybe a range for the coefficient?
What I did was carry out tan(theta) = Mu(s) plugged in 30 degrees for theta and got
Mu(s) = .577 and then said that the condition was that M(s) < .577 is that right?

Part C) Here I wasnt sure if I should use kinematics or conservation of energy equations. There's a friction force which is nonconservative, but I'm not sure how to incorporate this. Any help would be appreciated

 

[Dick]

Good job so far! Don't you want to say Mu(s)>.577? For part c) it starts with some KE. The PE cancels up and down. So you should be able to just subtract the frictional work from the initial KE to get the final KE, right? Frictional work is just frictional force times total distance travelled, also right?

 

[Return]

ok, sounds good, but how do I know the initial velocity to plug into my KE(i) equation?

 

[Dick]

2.4m/sec. Weren't you given that??

 

[alphysicist]

Hi Dick,

Good job so far! Don't you want to say Mu(s)>.577? For part c) it starts with some KE. The PE cancels up and down. So you should be able to just subtract the frictional work from the initial KE to get the final KE, right? Frictional work is just frictional force times total distance travelled, also right?

I think the original answer of \mu_s &lt; 0.577 is correct for this problem.

 

[Dick]

Hi Dick,



I think the original answer of \mu_s &lt; 0.577 is correct for this problem.

Oh, right. It's says "not to get stuck", not "stuck". Thanks for the sharp eye, alphysicist!

 

[Return]

oh ok, i think what I'm getting confused about is this: the sled goes up the ramp and then it comes down, can you assume the initial veloctiy coming down is 2.4 [m/s]? that was it's initial velocity when it was going up.Also can you also really disregard P.E.? because it goes up the ramp, stops, then starts going down the ramp. At the point it stops doesn't it have PE and not KE?

 

[Dick]

I was talking about considering the whole trip. Start at 2.4m/s. Go up. Stop. Come back down. In the round trip PE cancels. Guess I didn't make that very clear. You can also do what you suggest. Assume 0 KE at the top, compute the PE and subtract the frictional work to get the final KE.