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Exam question - difficult?

  • Thread starter lew0049
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  • #1
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Hey,
I am currently taking the pre-reqs for medical school and was wondering how difficult this question is for anyone interested. It was on our most recent exam and for Physics I, and I found this question to be too difficult considering there was another difficult conceptual problem (where you have to show your work) and then 10+ multiple choice questions. I reviewed numerous inclined plane problems - none were at all this difficult to understand.

Question: There is a hill 2.0 m high and 5.0 m in the base. A sled is sliding down the hill and stops after 35 m from the end of the hill. What is the friction coefficient?


We went over the answer in class which I vaguely understand. Essentially, I just wanted to see what you guys think of this question and how long it takes you to solve it (considering you are probably 100% better at physics that me). Thanks!
 

Answers and Replies

  • #2
Nabeshin
Science Advisor
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Well to me as a 2nd year undergraduate physics major, this requires essentially no thought (I don't come up with the number in my head instantly, but the equation is well formed). I remember this type of question though from AP physics or some other such class and would say it ranks in the medium category for that.

Do you want help understanding it?
 
  • #3
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Honestly, it would be greatly appreciated if you could help explain your thought process behind a problem such as this (because I'm sure this will type of problem will be on the final). The biggest problem I have with physics is having the correct mindset, if that makes sense. I have a B.S. and MBA w/ a finance concentration yet I struggle with this class more than any 5000 level finance class I've taken.
 
  • #4
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My teacher wrote out the formulas but its rather difficult to follow when you are trying to write down the formula also. I'm sure you know this but the final answer is u = h/(l+s)
 
  • #5
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My attempt included finding the x-values of forces but I had no clue what to do after I read the sled went off the inclined plane. So I was stuck with multiple equations yet no idea how how/where to plug them in. Thanks again.
 
  • #6
Nabeshin
Science Advisor
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Ok here's my breakdown...

I read the question: It's a dynamics (not equilibrium) problem and I notice two things. One, time is nowhere mentioned. Two, they're asking about friction. Both of these things instantly tip me off to use conservation of energy. With that in mind, you know that the fundamental equation is that energy is conserved. So PE+KE=PE+KE+FrictionWork. From the wording, PE and KE final are zero. PE initial is mgh, and KE initial is zero. Now just to find the form of the friction term. Work is force times displacement. Consider the hill portion. We slide down the hypotenuse, of length sqrt(2^2+5^2). The frictional force is mu times the normal force, which is mgcos(th) where th can be found by th=tan^-1(2/5). Then just add a term where the sled slides away from the hill, with coefficient mu and normal force mg. Once you have all this, it's a simple algebraic equation.

It looks like a lot when I write it out, but it's really not.
 
  • #7
rcgldr
Homework Helper
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Question: There is a hill 2.0 m high and 5.0 m in the base. A sled is sliding down the hill and stops after 35 m from the end of the hill. What is the friction coefficient?
Seems like insufficient information.

Is it a smoothly shaped bell curve like hill with a peak of 2.0 m, at 2.5 m from the edge of a base, or is it a slope instead that is 2 m tall and 5 m wide?

The initial condition of the sled wasn't stated either. Are we to assume the sled started at the top with zero initial speed?

Is the coefficient of friction on the hill the same as the coefficient of friction of the surface beyond the end of the hill?

Anyway, making the above assumptions, then you end up with two equations for coefficient of friction, one related to the velocity at the end of the slope during the trip down the slope, the other related to the distance it took to stop from that velocity on a flat surface.
 
Last edited:
  • #8
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Ok here's my breakdown...

I read the question: It's a dynamics (not equilibrium) problem and I notice two things. One, time is nowhere mentioned. Two, they're asking about friction. Both of these things instantly tip me off to use conservation of energy. With that in mind, you know that the fundamental equation is that energy is conserved. So PE+KE=PE+KE+FrictionWork. From the wording, PE and KE final are zero. PE initial is mgh, and KE initial is zero. Now just to find the form of the friction term. Work is force times displacement. Consider the hill portion. We slide down the hypotenuse, of length sqrt(2^2+5^2). The frictional force is mu times the normal force, which is mgcos(th) where th can be found by th=tan^-1(2/5). Then just add a term where the sled slides away from the hill, with coefficient mu and normal force mg. Once you have all this, it's a simple algebraic equation.

It looks like a lot when I write it out, but it's really not.

I really appreciate the input and I will respond in-depth tmw when my eyes aren't almost shut haha. The thought process is precisely the problem I have though.
 
  • #9
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Seems like insufficient information.

Is it a smoothly shaped bell curve like hill with a peak of 2.0 m, at 2.5 m from the edge of a base, or is it a slope instead that is 2 m tall and 5 m wide?

The initial condition of the sled wasn't stated either. Are we to assume the sled started at the top with zero initial speed?

Is the coefficient of friction on the hill the same as the coefficient of friction of the surface beyond the end of the hill?

Anyway, making the above assumptions, then you end up with two equations for coefficient of friction, one related to the velocity at the end of the slope during the trip down the slope, the other related to the distance it took to stop from that velocity on a flat surface.
Exactly, without some sort of velocity or acceleration given I found this problem to be very confusing. Here is how our teacher broke it down (and no additionally information was given regarding the above assumptions and no diagram was given):

Note: when she worked the problem and showed a picture, L is the horizontal length of the system, point B was the when the incline ended, and point C was where the sled stopped (35m), and point A was the top of the incline

PE(a) = WF(f)a->c = W(ab) + W(bc)
F(ab) = u*mg cos theta
F(bc) = u*mg
W= u*mg cos theta*|AB| + u*mg*|BC|
cos theta = L / |AB|
W(ab) = u*mg (L / |AB|*|AB|
W = u*mg*L+u*mg*s = m*g*h
u*(L+s) = h
u = h / (L+s)

I believe those are in order but what confused me is that I kept trying to find some form of an acceleration or velocity, specifically acceleration. It would have just be nice to see a problem somewhat similar to this but instead we've never done any type of problem involving an inclined plane + level ground friction problem. It's the initial setup that got me
 
  • #10
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I'm assuming the sled started from rest.
 
  • #11
rcgldr
Homework Helper
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PE(a) = WF(f)a->c = W(ab) + W(bc)
F(ab) = u*mg cos theta
F(bc) = u*mg
W= u*mg cos theta*|AB| + u*mg*|BC|
cos theta = L / |AB|
W(ab) = u*mg (L / |AB|)*|AB|
W = u*mg*L+u*mg*s = m*g*h
u*(L+s) = h
u = h / (L+s)

I believe those are in order but what confused me is that I kept trying to find some form of an acceleration or velocity
If you had tried to determine the velocity, or the KE related to velocity 1/2 m v^2, then you would have ended up with:

1/2 m v^2 = m g h - u m g L
1/2 m v^2 = u m g s

subtacting the 2nd eq from the 1st

0 = m g h - u m g L - u m g s

u m g s + u m g L = m g h
 

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