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Work and Energy

  1. Mar 5, 2008 #1
    [SOLVED] Work and Energy

    Hello, I'm a little new to this forum but I would really appreciate some help with this particular problem.

    A roller coaster reaches the top of the steepest hill with a speed of 5.8 km/h. It then descends the hill, which is at an average angle of 25 degrees and is 50m long. What will the speed be when it reaches the bottom. assume the coefficient of friction (uk) = 0.12

    So the first thing I did was convert the 5.8 km/h to 1.61 m/s to get consistent units. I know that KE initial + PE initial = KE final + PE final. I also know I have to break things up into components. I know Fx=mgcos25. But I think I'm screwing up the Fy component. After that I'm lost and I'm not sure where to go from there. Any suggestions and help will be greatly appreciated.
  2. jcsd
  3. Mar 5, 2008 #2

    Shooting Star

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    What do you mean by Fx here?
  4. Mar 5, 2008 #3

    Doc Al

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    Staff: Mentor

    What about the work done by friction?
    That's the component of the weight normal to the surface. What's the component of the weight parallel to the surface?

    What's the friction force? The work done by friction?
  5. Mar 5, 2008 #4


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    Calculate the forces first

    Hi peaches! Welcome to PF! :smile:

    Hint: Be systematic.

    1. How many forces are there?

    2. What is the value of each force?

    Then you can start deciding what to do. :smile:

    (btw, if you type alt=m, it prints µ for you!)
  6. Mar 5, 2008 #5
    thank you all for your responses. I looked back at this problem and looked again at the forces affecting the object, the normal force, the weight, and friction. Please tell me if I'm going in the right track. If I find the net work, that would be delta KE. Then I would just have to solve for the final velocity. Does that make sense?

    I know that force of friction is µkFN.
    X: mgsin(theta)= ma
    Y: FN - mgcos(theta) = 0
    thus, FN = mgcos(theta)

    the work of friction when simplified is equal to µkmgcos(theta)dcos(180)
    there is no work by the normal force cos(90)= 0

    is there work done by the weight??
    Last edited: Mar 5, 2008
  7. Mar 6, 2008 #6

    Shooting Star

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    Go back to the start, when you had written initial KE+PE = final KE+PE. That would have been correct if there were no frictional force.

    The object has to to do work against the frictional force when coming down the incline. If F is the frictional force, then work done against friction is F*distance.

    So, initial (KE+PE) = (final KE+PE) + F*d.

    You know F=µmgcosθ. Now you should be able to solve directly.
  8. Mar 6, 2008 #7

    Doc Al

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    That's a perfectly valid approach as long as you make sure you deal with the work done by every force.

    The net force equals ma (you forgot the friction).
    Good. Note that the work done by friction is negative.

    Most definitely. What's the component of the weight along the direction of motion?

    Note: While this is a perfectly valid approach, it requires that you account for the work done by gravity. If you use the "energy method" that you started with (and as explained by Shooting Star in post #6), that work is automatically taken care of as a PE term.

    It might be easier to understand the energy method if you write it as:
    initial (KE + PE) + Work done by friction = final (KE + PE)

    Since the work done by friction is negative (= -F*d), that becomes:
    initial (KE + PE) -F*d = final (KE + PE)

    (which is equivalent to what SS wrote, of course)
    Last edited: Mar 6, 2008
  9. Mar 6, 2008 #8
    thank you all for your help!
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