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Power output by basketball player when jumping

  1. Nov 18, 2014 #1
    1. The problem statement, all variables and given/known data

    A ##\small{105\!-\!kg}## basketball player crouches down ##\small{0.400\ m}## while waiting to jump. After exerting a force on the floor through this ##\small{0.400\ m}##, his feet leave the floor and his center of gravity rises ##\small{0.950\ m}## above its normal standing erect position.

    What was his power output during the acceleration phase?

    ##m = 105\ kg##

    ##d = 0.400\ m##

    ##h = 0.950\ m##


    2. Relevant equations

    ##P = \frac{W}{t}##

    ##\Delta PE_g = \Delta KE##

    3. The attempt at a solution

    The problem disregards the work done by the player when first crouches, so the work I tried to find is just when he accelerates upwards (jumps).

    First we know that: ##P_{output} = \frac{W_{output}}{t} = \frac{F_{exerted}d}{t}##

    The velocity with which he left the floor is the same to that just before he touches the floor when landing; and after having fallen a height ##h##:

    ##\Delta PE_g = \Delta KE##

    ##mgh = \frac{1}{2}mv^2##

    ##v= \sqrt{2gh} = \sqrt{(2)(9.8\ m/s^2)(0.950\ m)} = 4.32\ m/s##


    Assuming that the velocity was obtained after a net constant acceleration along a distance ##d##, then:

    ##a = \frac{v^2}{2d} = \frac{2gh}{2d} = \frac{(9.8\ m/s^2)(0.950\ m)}{0.400\ m} = 23.3\ m/s^2##


    Then, I try to find the average force exerted:

    ##F_{exerted} = F_{net} + w##

    ##F_{exerted} = m(a + g)##

    ##F_{exerted} = (105\ kg)[23.3\ m/s^2 + 9.8\ m/s^2] = 3473\ N##


    Then, I tried to find the time:

    ##t = \frac{v-v_0}{a} = \frac{4.32\ m/s}{23.3\ m/s^2} = 0.185\ s##


    And finally the power:

    ##P_{output} = \frac{F_{exerted}d}{t} = \frac{(3473\ N)(0.400\ m)}{0.185\ s} = 7.49\ kW##

    But the answer provided by the textbook says that the power must be ##8.93\ kW##. After playing around with the numbers, I noticed that what makes the difference between both results is the way the time was calculated. According to the textbook, the time is:

    ##t= \sqrt{\frac{2d}{(a + w)}} = \sqrt{\frac{2(0.400\ m)}{(23.3\ m/s^2 + 9.8\ m/s^2)}} = 0.156\ s##

    My question is: why should ##g## be added to the ##a##? It's like considering an absolute acceleration exerted by the player, but that should underestimate the time the force was exerted.

    In a hypothetical situation, if I were pushing a box with an acceleration of ##1\ m/s## through a distance ##d##, and it happens that I know friction is exerting an acceleration of ##3\ m/s## in the opposite direction, the time I spent pushing the box is dependent of ##a## but not ##a + a_{friction}##. The latter will yield a shorter time than that I really spent pushing the box (and hence, a greater power output).

    I'm puzzled.
     
  2. jcsd
  3. Nov 19, 2014 #2

    NascentOxygen

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    Hi Jazz. You have presented your case well. I agree with you that the textbook solution is wrong.
     
    Last edited: Nov 19, 2014
  4. Nov 19, 2014 #3

    haruspex

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    The question does not make it clear that the force is constant, while the power to be determined is an average. If that is correct I agree with your answer.
    More realistically for muscle, power is constant. But that leads to quite a messy equation, with no solution in closed form.
     
  5. Nov 19, 2014 #4
    Thank you, both, for your answers.

    The problems involving force that I've been dealing with so far just consider average acceleration. I know that this does not happen in real life (at least not in situations as in the problem) but I get very confused trying to understand how a problem with changing acceleration must be treated (as far as I understand, with help of calculus I will be able to do so).

    In this case, if I were asked to find a constant power output, does it mean that the magnitude of the force gets reduced, but the time over which it acts gets longer along the distance ##d##, 'fitting' an equation like this ##\frac{\Delta F}{\Delta t}d##?

    If there is no closed form for calculating constant power, is it calculated by taking shorter distances ##d## along which a force ##F## acts during a time ##t## and then add them, or by taking shorter times over which it acts and add them?

    If I were asked to find
     
    Last edited: Nov 19, 2014
  6. Nov 19, 2014 #5

    haruspex

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    Not sure you've understood what I mean by 'no closed form'. Using calculus, you get an equation for P which involves P in 'incompatible' ways. Specifically, I got ##mg(d+h) = -P\sqrt{\frac{2h}{g}} - \frac{P^2}{mg^2}\ln\left(1-\frac {mg}{P}\sqrt{2gh}\right)## , but that's so horrible I suspect I made an error. At least it's dimensionally correct.
     
  7. Nov 19, 2014 #6
    Roughly speaking, I understood a closed form as a nice formula used to solve 'something' that seems to be messy. Maybe it's too rough.

    I hope to get into the math soon to understand how to get the equation you got. Just for curiosity, ##mg(d+h)## is considering the distance he crouches, why is that?

    So when considering constant P, both I and the textbook are wrong, right?
     
  8. Nov 19, 2014 #7

    haruspex

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    That's the total work done, no?
    Without completely solving it, I can't be certain it won't give the textbook answer, but it seems unlikely. As I wrote, my equation could be wrong.
     
  9. Nov 19, 2014 #8
    Yep, I was just overthinking about it.

    (:

    Thanks for helping me to see the light with this.
     
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