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Calculating Work/Power

  1. Nov 19, 2009 #1
    Since Work= F displacement cos theta. If a person runs up a flight of stairs and then back with a constant velocity, does that mean his acceleration is also 0. If that's the case is the person's force 0, making his work also 0. This doens't make sense to me. Or is acceleration, one due to gravity, so it would be 9.8 m/s^2? Does this change when a person goes down a flight of stairs or is the same?
  2. jcsd
  3. Nov 19, 2009 #2
    The work done against gravity is mgh but the acceleration of the person is whatever it is as they are going up and down the stairs. Remember they need to keep applying a force and therefore an acceleration to keep moving up the stairs. The average velocity is somewhat constant but what that is more depends on the power P = W/t = FV the rate at which the work is done( how quickly they get up the stairs.) But the certainly have to a apply an acceleration to maintain this velocity. In short bursts going up the acceleration has to exceed 9.8 for the runner to get anywhere but coming down it would generally be small chucks of 9.8m/s^2 down as they step down each step and are stopped by the floor. Otherwise they come crashing down at some ridiculous velocity!
  4. Nov 19, 2009 #3
    When going up the stairs, your body is always working against gravity, so even though you are going at a constant speed you are still applying a force up against gravity to cancel it out, hence no acceleration. But if you go up then down the stairs at a constant speed, your height will be the same as before, therefore you would have moved no where and your work is 0. And just to clarify, Work= force*displacement(both of these are vectors) meaning the force in the direction of the displacement. since you are not accelerating up the stairs then the only direction you are using a force is up against gravity.

    Does this make sense?
    Sincerely FC
  5. Nov 19, 2009 #4
    1. you are going at a constant velocity therefore there is no change in velocity, and since acceleration is the rate of change in velocity, acceleration=0
    2. no, acceleration due to gravity does not change going up or down. the only thing that has changed is that instead of you pushing against gravity when you are going up, when your going down gravity is pushing you
  6. Nov 19, 2009 #5
    Your responses contradict each other.

    Is there an acceleration or not?

    @FoxCommander- Are you saying that my displacement would be 0? How is it zero if I did not end up at the same spot as I started? Why wouldn't the displacement be the distance between the ground (starting point) and the top of the stairs?
  7. Nov 19, 2009 #6
    ok sorry for the confusion ha ha, lets just simulate the experiment.
    so you are at the bottom of the stairs. the stairs are at some angle theta from the ground. the top of the stairs have some hypotenuse 'd'
    you then proceed to walk/run whatever up the stairs at a constant speed till you reach the top and stop.
    If you have gone the entire way at a constant speed then your acceleration would be zero
    however, gravity is in this case a constant force with acceleration 9.8m/s2 and you must overcome this force to reach the top of the stairs. since it is always pointing down(towards the center of the planet) then you must also push an equal force in the up direction. and so we then go to the equation W=F(vector)times d(also a vector). The distance traveled in the Y-axis is equivilant to the Hypotenuse*sin(theta) <assuming that your angle is from the ground> and so you say the F(equivilant the the force of gravity on you,ie your weight) multiplied by the height of the stairs is the work you have done

    As for the part were the work is zero, if you say climb down the stairs then it is not you who is pushing but gravity and so the work would be negative. And since you go the same vertical distance up and down then you will have a total of zero work done

    Again really sorry about the confusion, first time forum-er
    Sincerely, FC
  8. Nov 19, 2009 #7
    I really appreciate your help, but I'm really slow, so I apologize.

    So what you are saying is that I worked to get up the stairs, but when I go down my work becomes 0?

    Oh and Iwhat I meant was that your post and lubuntu's post contradicted each other because she said there was an acceleration. Sorry for the confusion on my part.
  9. Nov 20, 2009 #8
    no no its fine, i enjoy helping ha ha.

    well now lets isolate just the coming down part. If you think about it, when you are coming down are you doing anything really? all you are doing is letting gravity 'push' you down each step. so technically speaking you are actually doing negative work in the positive y-axis because you are going down..... if that makes sense. Im a much better explainer visually ha ha. so by the time you have gone down the stair you have done the same magnitude of work as you did going up only it is negative because the force is in the opposite direction. So if you went both up and down and so you would say Work up+Work down= Work up-Work up(since they are both equal)= 0
    Disregard the whole Acceleration thing thats just me saying nonsense ha ha
  10. Nov 20, 2009 #9
    I understand what you are saying. Let me try using numbers to see if I really understand it.

    w= F delta x cos0

    a= 9.8 m/s^2
    m= 50 kg
    delta x= 3m

    W= (9.8 x 50) (3) cos 0
    W= 1470 J

    w= F delta x cos0

    a= -9.8 m/s^2
    m= 50 kg
    delta x= 3m

    W= (-9.8 x 50) (3) cos 0
    W= -1470 J

    Can work be a negative number though?

    1470J-1470 J = 0 J of work done
  11. Nov 20, 2009 #10
    and yes it can be because force is a vector, i dont know if you know what that means but basically it means it has a magnitude and a direction in which it is pointing. so if the direction is up and you go down, then the displacement will be negative,
    Another way to think of this problem is to think of it as stored Potential Energy. If you go up the stairs you are storing energy in the form of Potential Energy due to the change in height. but when you go down you are losing that potential energy until you reach the bottom where the height is zero and so W=MGH ,M-mass, G-accel due to gravity, H-height, and your height is zero so the whole thing is zero
  12. Nov 20, 2009 #11
    I forgot that it was a vector. xD

    Is this how you would compute that though? Are my calculations correct?
  13. Nov 20, 2009 #12
    yes, the only problem i see is that you use cosine, make sure you know the SOHCAHTOA rules when using cosines and sines. Not that you cant use cosine for this but it depends on which angle of the triangle you are using
  14. Nov 20, 2009 #13
    I think I understand it now. Thank you very much for being so patient with me. :]
  15. Nov 20, 2009 #14
    No problem mr./mrs. I really enjoy teaching others physics If you ever need anything else let me know, you know my name :)
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