TBR Physics Example 3.8b (Work and Energy)

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The question in this example (page 106, example 3.8b on work and energy) is:

A bicyclist skids to a complete stop under three different scenarios
I. on a dry flat road while travelling at 10.17 m/s (uk = 0.72)
II. on a wet flat road while travelling at 10.36 m/s (uk = 0.60)
III. On an oily road with a slight downward slope while travelling at 9.87 m/s (uk = 0.37)

What can be said when comparing the three scenarios?

A. Less work is done by the road on the bicyclist in Scenario I than in Scenario II
B. Less work is done by the road on the bicyclist in Scenario II than in Scenario III
C. The amount of kinetic energy lost by the bicyclist in Scenario III depends on the angle of the downward slope
D. The amount of work done on the bicyclist in Scenario III depends on the angle of the downward slope.
Answer: D



The answer key states that D is the correct answer - I've attached the solution explanation to this post as well. While I understand why D is correct, I don't fully understand why option A is incorrect. I would have thought that less work is done by the road on the bicyclist in Scenario I than in Scenario II, since vi (and so change in KE = W, since vf=0 in all three scenarios) is lower in Scenario I (10.71 m/s) than in Scenario II (10.36 m/s).

Re-reading the solution explanation, it seems to conflict directly with the question (ex: "the bicyclist is going fastest in scenario I") that's left me pretty confused. Any help would be appreciated!

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The answer key states that D is the correct answer - I've attached the solution explanation to this post as well. While I understand why D is correct, I don't fully understand why option A is incorrect. I would have thought that less work is done by the road on the bicyclist in Scenario I than in Scenario II, since vi (and so change in KE = W, since vf=0 in all three scenarios) is lower in Scenario I (10.71 m/s) than in Scenario II (10.36 m/s).

There is also a larger coefficient of kinetic friction on the dry road as compared to the wet road. Remember work isn't a state function and the work-energy theorem only applies when you have all conservative forces acting on the object. When you have non-conservative forces acting, it's total energy that is conserved and that includes the heat energy dissipated by the friction on the tires.

But also note that there really isn't such a thing as work done by friction since friction is by nature a dissipative force and does not really act in any one direction on a macroscale. We can only characterize it on the average macroscopically so that we can model systems but friction is a force that acts at the microscopic level, on particles.
 
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There is also a larger coefficient of kinetic friction on the dry road as compared to the wet road. Remember work isn't a state function and the work-energy theorem only applies when you have all conservative forces acting on the object. When you have non-conservative forces acting, it's total energy that is conserved and that includes the heat energy dissipated by the friction on the tires.

But also note that there really isn't such a thing as work done by friction since friction is by nature a dissipative force and does not really act in any one direction on a macroscale. We can only characterize it on the average macroscopically so that we can model systems but friction is a force that acts at the microscopic level, on particles.


Thanks for your reply! Perhaps i'm misunderstanding, but I had thought that as Wnet = deltaKE and Wnet = Wnc + Wc, if no non-conservative forces are present (Wnc = 0), Wnet = Wc = -deltaPE, so we can use the law of conservation of energy in those cases (KEi + PEi = KEf + PEf), but not if non-conservative forces are present, as that expression would not hold true.

However based on the definition of work that's used to derive the law of conservation of energy in the first place, the work energy theorem (Wnet = deltaKE) would have to apply equally to scenarios where there are both conservative and non conservative forces as well. Doing some googling seems to be support this here, as well as in this textbook (and had been what I had assumed, as this was stated repeatedly when I took General Physics as well). How would Wnet = deltaKE be invalidated by the presence of non-conservative forces?

As gravitational potential energy doesn't impact the work done in scenarios I and II (although it does in scenario III) and vf = 0 in both scenarios I and II, W = deltaKE = deltaKEi = W from friction, which in scenario I is -½m(10.17)^2, but -½m(10.36)^2 in scenario II that will always be larger (and would make statement 1 true). As distance traveled is not the same between scenarios I and II, and the bicyclist always travels until he/she comes to stop, the information regarding friction coefficients did not seem to contradict this (as the larger coefficient in scenario would be counteracted by travelling a longer distance).
 
However based on the definition of work that's used to derive the law of conservation of energy in the first place, the work energy theorem (Wnet = deltaKE) would have to apply equally to scenarios where there are both conservative and non conservative forces as well. Doing some googling seems to be support this here, as well as in this textbook (and had been what I had assumed, as this was stated repeatedly when I took General Physics as well). How would Wnet = deltaKE be invalidated by the presence of non-conservative forces?

Yes, I think you are right actually. I'm a chemist by training so I'm no physicist but now that I think about it more, it makes sense intuitively. Since the car had some kinetic energy before and now it doesn't and so the work done to effect that change in kinetic energy must be frictional (that is, if we assume that frictional work is a thing - there are good reasons as to why it's not). So if you ignore the fact that frictional work isn't really a thing, I think your argument makes sense.
 
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Can't envision any justification for the speed being recited as greater in Scenario I. Looks like a goof.
 
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