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Motion on an inclined plane

  1. Jul 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A car is moving on a horizontal road with speed of 60 km/h. When it starts to move on an inclined plane its velocity decreases by 10 km/h. How much its velocity will increase when the car starts to move down the plane? Assume that the power of the engine remain the same, any friction is neglectable.

    2. Relevant equations
    Newton's Laws of Motion

    3. The attempt at a solution

    The correct answer is 15 km\h. I don't understand this. Please help.
  2. jcsd
  3. Jul 28, 2015 #2
    Insufficient info and a poorly posed question.
  4. Jul 28, 2015 #3
    I've rechecked the problem. There is nothing more given. Note that it's a problem from my country's physics olympiad so I expect it to be difficult.

    Maybe the way I'm trying to solve it is wrong, but I just don't know where to start.
  5. Jul 28, 2015 #4
    In a frictionless world, how much power is required to maintain 60 km/h horizontally?
  6. Jul 28, 2015 #5


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    I think the phrase "neglect any friction" is misleading. For the problem to work out, you have to assume that there is "internal friction" that the engine must overcome even when driving on a horizontal road. So, the engine has to provide a certain power while driving horizontally. I believe the problem wants you to assume this internal friction is the same even when the car is going up or down hill and that this is the only friction you need to worry about.

    Think about the forces acting on the car while going uphill and downhill.
  7. Jul 28, 2015 #6
    Well in the statement of the problem it is said that air fiction is neglectable but since there is nothing said about other kind of frictions, I thought that any friction is neglectable.
  8. Jul 28, 2015 #7


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    If there is no friction internally or externally, then the engine would not need to provide any force to keep the car moving on a horizontal surface. There must also be static friction between the tires and the road when going uphill.

    I think you have to assume that the engine provides a force Fengine,H while moving horizontally. So the engine is providing a certain power while moving horizontally. Then, when the car is going uphill, the engine must provide a greater force Fengine,UP. But you are supposed to assume that the engine still provides the same power while going uphill (or downhill).

    Symbolically, can you express the force Fengine,UP in terms of Fengine,H and some other force?
    Likewise, can you express the force Fengine,DOWN in terms of Fengine,H and the same "other force"?
  9. Jul 28, 2015 #8


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    I agree that the wording of the problem is open to interpretation.

    But it appears that you will get the correct answer if you look at it as follows. As you wrote, the power of the engine while moving horizontally can be expressed as PH = FHvH. And the power of the engine while moving uphill can be expressed as P1 = F1v1. You need to think about why F1 is greater than FH and how you can express F1 in terms of FH and some other force.
  10. Jul 28, 2015 #9


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    I believe your mistake is in the equation attached below. How did you get this? Shouldn't F2 be less than FH? The engine will not need to provide as much force coming down the hill compared to traveling horizontally.

    Attached Files:

  11. Jul 28, 2015 #10
    Because of P1=P2 I got a ratio FH=0,8333F1. It means that horizontal force is 16,7% bigger than the force with which the car is moving upwards. Then I made the assumption that the force with which the car will move downwards should be bigger than the horizontal force by the same amount of percent. I see that my answer is somewhat unreasonable.

    You say that F1 should be greater than FH but how about PH=PUP? It gives me a clear ratio FH=0,8333F1.

    Hovever, I've found another solution:
    It's still not the right answer but close.
  12. Jul 28, 2015 #11


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    But doesn't FH=0,8333F1 imply that FH is smaller than F1. So, F1 > FH.
  13. Jul 28, 2015 #12


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    First, I don't see a need to write FH as μmg. I would just leave it as FH.

    For your expressions for FUP and FD, I believe you have some sign errors. FUP should be greater than FH in order to overcome the effect of gravity when going uphill.
  14. Jul 28, 2015 #13
    Does this mean the final answer should be 50+15= 65 km/h or 60 +15= 75 km/h?
  15. Jul 28, 2015 #14


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    I believe it should be 75 km/hr. This is another ambiguity in the wording of the problem statement. :confused:
  16. Jul 28, 2015 #15
    I agree. (Only because I finally was able to get that answer.)
  17. Jul 29, 2015 #16
    Everything turned out fine.
    The answer is 15 km/h because the problem is asking how much it will increase, therefore Δv=75-60=15 km/h.
  18. Jul 29, 2015 #17
    Well done, my friend. Treating FH as a simple friction force leads to a clear understanding of how the forces interact.
  19. Nov 2, 2015 #18

    I have a doubt regarding the concept of work/energy/power when it comes to car fuel problems .

    Isn't the force provided by engine an internal force ? Because if that is true then how does it appear in force balance equations ?

    This is what I understand . When a car moves on a road without slipping then there is no net work done on the car . Yet the car accelerates . The internal(chemical) energy of the fuel gets converted in kinetic energy .

    But I fail to see how is force provided by engine an external force .

    Please help me remove this doubt .
  20. Nov 3, 2015 #19


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    Hello, Vibhor. The term "force provided by the engine" that I used is vague and I probably should have avoided it. In order to make sense of the problem, I assumed that the engine must expend power just to keep the car moving horizontally at constant speed. The reason the engine would need to supply power for horizontal motion might be due to internal friction (in the axles, for example) that would dissipate the translational kinetic energy of the car if the engine was not supplying power. I assumed that the rate at which energy is dissipated internally is proportional to the speed of the car. Thus, you could write this power expended by the engine for horizontal motion as PH = FHVH where FH has the dimensions of force and is assumed to be a constant.

    When going uphill, the engine must supply additional power to account for the rate of increase of gravitational potential energy of the car. So, the total power supplied by the engine while going uphill would be Pup = FHVup + mgsinθVup. Here, it is assumed that FH associated with the internal dissipation of energy has the same value as for horizontal motion.

    For going downhill, Pdown = FHVdown - mgsinθVdown.
    The problem can be solved by setting PH = Pup = Pdown (as stated in the problem).

    I'm not sure if this was how the problem was intended to be interpreted.

    I agree. It's like a person standing on a frictionless floor facing a wall. By pushing on the wall, the person can start moving away from the wall. But, the force of the wall on the person does not do any work. The kinetic energy gained by the person comes from conversion of energy stored inside the person into kinetic energy of the person.

    I agree with you that the "force provided by the engine" is an internal force, not an external force. The only external forces acting on the car are the normal force from the road, friction force from the road, the force of gravity, and possibly an external drag force (which I took to be zero in the problem). When the car is traveling at constant speed horizontally with no external drag, then the friction force from the road is zero.
  21. Nov 3, 2015 #20
    Nice explanation !

    Would you agree that in order to accelerate the car the fuel input rate to the engine would have to be increased ? This would increase the force exerted by engine on the tyres (rather on the piston) which in turn would increase the friction on the tyres and hence the acceleration ?

    This raises another question -

    Does higher rotation rate of the wheels mean more slipping tendency resulting in increased friction force ?
    Last edited: Nov 3, 2015
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