Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Power output physics problem

  1. Sep 17, 2005 #1
    heres this question:
    theres a man who does 20 press-ups (ie facing down body straight) in 50 sec what is his power output?

    is it 20/50 = 0.4 press-ups per sec of power?

    i know power is in watts and its P = Fv = E/t

    if i take accleration to be 9.8 i could get v, a= (v-u)/t
    initial vel would be 0 ithink but i still can't get force i require mass which im not given in the Q so should i assume a mass?
    or is the answer above correct? and im thinking too much into it?
     
  2. jcsd
  3. Sep 17, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You'll need to estimate how much work the man does during each press-up. He is lifting some percentage of his bodyweight through some distance.
     
  4. Sep 17, 2005 #3
    isnt he lifting his whole body so 100%of his body weight so force would be 9.8 * (100% of mass) =
    i could use S = 1/2 at^2 to find displacement? i tried that and i get a silly answer of 7. summin meters (thats using t as (50/20)/2 and a as 9.8 initial vel as 0)
    and say i could get vel by a= (v-u)/t

    then mav= mas/t = P
    help how do i find some % of bodyweight and some dist.
     
    Last edited: Sep 17, 2005
  5. Sep 17, 2005 #4

    Doc Al

    User Avatar

    Staff: Mentor

    What matters is how much force he exerts. An example: What's easier, lifting one end of a sofa off the ground or lifting the entire sofa?

    Why not check this yourself? Find a bathroom scale. Stand on it to get your weight. Then do a pushup on it, and see what it reads.

    Those are formulas for freely falling bodies. I don't see anything falling here. The distance? Hint: How long are your arms?
     
  6. Sep 17, 2005 #5
    m*9.8*12.25 = m*9.8*7.66 but the mass cancels and the numbers are silly no one got arms 7 meters long
     
  7. Sep 17, 2005 #6
    my arms are very short lol. so are u suggesting i should assume values? like the average persons weight and arm length?etc
     
  8. Sep 17, 2005 #7
    hang on my arm length is about a third of the length of my body or summin should i be thinking along those lines? ok maybe not a third. ( ive got a new teacher for physics, i dont want him to think im dumb or something)
     
    Last edited: Sep 17, 2005
  9. Sep 17, 2005 #8

    Doc Al

    User Avatar

    Staff: Mentor

    If you're not sure how far you press yourself up, get a ruler and check. All you need is an estimate. (Did you check what percentage of your weight is involved in a press up? Do it!)

    This problem is just an exercise to see if you understand the ideas of work and power. All you will count, I presume, is the positive work done as the man pushes himself up. You can use your own weight/force/distance as an estimate, or you can express it in terms of the unknown man's weight and arm length. (But don't start talking about falling objects! :wink: )
     
  10. Sep 17, 2005 #9
    ok :blushing: thanks i think i did look too much into the question.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook