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Power developed by a jumping man

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A 160-lb man leaps into the air from a crouching position. His center of gravity rises 1.5ft before he leaves the ground, and it then rises 3ft to the top of his leap. What power does he develop assuming that he pushes the ground with constant force?

    Ans. clue: More than 1hp, less than 10hp

    2. Relevant equations
    Power

    3. The attempt at a solution

    I have trouble understanding this problem but made an attempt at a solution.
    Call h1 = 1.5ft = 0.46 m, and h2 = 3ft = 0.92m, m = 160 * 0.45 = 72kg

    As far as I understand, the power the man develops is the work done on the ground divided by the time his feet are in contact with the ground.
    The working forces on the ground are the weight downward, and the ground reaction force upward. Because it is assumed constant, the ground reaction force is conservative and has a potential function which is -N x height.
    So the power developped is

    ## P = \frac{W_g}{\triangle t} = - \frac{\triangle U_g}{\triangle t} = \frac{N-mg}{\triangle t} h_1##

    Finding the time interval :
    Because all forces are constant, it is easy to find ##\triangle t## with the impulse-momentum relationship:

    ## (N-mg) \triangle t = m v_{jump} - 0 = m v_{jump}##

    The jumping speed is found by using conservation of energy once the man is in the air, and:

    ##{\triangle t} = \frac{m}{N-mg}\sqrt{2g(h_2-h_1)} ##

    Finding the ground force :
    Once again, I use conservation of energy, but this time on the ground:
    ##E_{jump} = E_i = 0 \Rightarrow N = \frac{1}{2h_1}m{v}_{jump}^2 + mg = mg\frac{h_2}{h_1}##


    All together, the power is :

    ## P = \frac{m}{\sqrt{2} h_1} (g(h_2-h_1))^{3/2} \approx 1.42 hp##

    How ugly is this ?
     
  2. jcsd
  3. Jan 20, 2015 #2

    mfb

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    The force is constant, but the speed is not, so power will vary over time.

    I don't understand how you calculated the ground force, but the result looks wrong (just an issue with the last step I think).
     
  4. Jan 20, 2015 #3
    Hello, thanks for the reply.
    I don't really understand the question in the problem: what does it mean to 'develop a power' ?

    My first thought was that I had to find the average power during the time the man had its feet on the ground.
    Because the forces are conservative, it can be written as a difference of potential energy during that time. The potential energy during that time is ## U(y) = (mg - N) y ##.

    How I calculated N :
    I used conservation of mechanical energy. At the beginning the man is at crouching position at rest, so ##E_0 = 0##. At height ##h_1##, as his feet leave the ground, ##E_{h_1} = \frac{1}{2} m {v}_{jump}^2 + U(h_1) ##.

    But I'm not sure :) What do you think ?
     
  5. Jan 20, 2015 #4

    mfb

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    Power the human needs to accelerate.
    The problem statement does not make clear if average or maximum power is asked for, it is probably safe to give both (once you have one, the other one is not so complicated).

    Okay. How did the last "=" sign work? This one:
     
  6. Jan 20, 2015 #5
    Okay, last part is obtained by replacing ##{v}_{jump}^2 ## by its value, which is given by conservation of mechanical energy between the time the man leaves the ground, and the time he is at the top of his leap. Here, only the weight is working, so ##{v}_{jump}^2 = 2g(h_2-h_1) ##.

    Let me think for maximum power ...
     
  7. Jan 20, 2015 #6

    mfb

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    As far as I understand the problem statement, h2=.92m is the height difference between the point where the man leaves the ground and the highest point of the jump, then you should not subtract h1.
     
  8. Jan 20, 2015 #7
    yes I was in doubt about that too, I also tried h2 = 4.5 ft which gives an average power of 4.02 hp.
     
  9. Jan 20, 2015 #8
    I think maximum power occurs as the man leaves the ground because power is a strictly increasing function relative to time:

    ## \frac{dP}{dt} = (N-mg) \frac{dv}{dt} = \frac{(N-mg)^2}{m} > 0 ##

    So ## P_{max} = (N-mg) v_{jump} = (N-mg) \sqrt{2g(h_2 - h_1)} ##
     
  10. Jan 20, 2015 #9

    mfb

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    Right.
    I don't see why you subtract mg from the force, it is still a force the legs have to exert.
     
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