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Homework Help: Work and Power

  1. Aug 16, 2006 #1
    Hey everyone. I'm new to this site and was wondering if u can help me with a work and power problem. :smile:

    A 60kg student runs at a constant velocity up the incline described in the diagram in 4.5s. Calculate the power output of the student.

    The diagram was a picture of a right-angled triangle with hypotenuse length 4 meters. The hypotenuse is the ramp, and the ramp is raised to a height of 2.5m (also the length of the vertical leg). The student is running up the ramp.

    My Work:
    Since he is running at constant velocity, acceleration = 0m/s/s.
    F=ma, so F = 0 N.
    W=Fd, so W = 0 J.
    P=W/t, so P = 0 W.

    The answer from the back of the book was P = 326.67 W.

    I know the student is doing work in terms of his muscles having a tensional force, but I don't think that is the way to go with this problem at all.

    Anyways, any help will be appreciated thanks.

    BTW, there was no information regarding friction.
  2. jcsd
  3. Aug 16, 2006 #2


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    Homework Helper

    The student has to overcome some gravitational pull, thus he has to generate force. Can you determine its magnitude?
  4. Aug 16, 2006 #3
    The fact that the student is doing work is pivotal to finding the power. It's easier to think of this in terms of energy. The work that the student does brings him from the bottom of the incline to the top of the incline. What type of energy (kinetic or potential) changed for the student from the bottom of the cliff to the top of the cliff? Hint: The student is assumed to keep a constant velocity throughout.

    What you did was assume that since the student is not accelerating, he is doing no work. But does it really make sense to think that someone can get from the bottom of a cliff to the top of the cliff without doing any work? (I made the same mistake plenty of times, so don't feel bad.)

    The student's work creates a change in the student's energy, bringing him from the bottom of the incline to the top of the incline. If the work is W, then, as you already found, power is work over time.
  5. Aug 16, 2006 #4


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    Staff: Mentor

    Another way to approach this kind of problem is to write the total energy as the sum of the potential and kinetic energies:

    TE = PE + KE

    If his velocity is constant, are there any changes in his KE as he runs up the hill? What can you say about his PE? And then just remember that power is the amount of energy change per unit time (per second in mks units).

    EDIT -- Saketh was too fast for me -- beat me to it.
  6. Aug 16, 2006 #5
    Thanks for the help, I solved the problem by thinking Power as change in energy over time.
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