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Homework Help: Height of stairs

  1. May 25, 2010 #1
    A 65.0 kg student climbs a set of stairs in 25.0 s with a power output of 95 W. What is the vertical height of the stairs?

    I came up with the formula:

    W=P x (delta t)
    W=95w(25.0s)
    2375

    Im not sure how this comes in to play:
    W = mg (delta h)


    Im really confused :|
     
  2. jcsd
  3. May 25, 2010 #2

    collinsmark

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    You're on the right track. :approve:

    Work is equal to force times distance. mg is the force. delta h is the distance.

    [Edit: Technically, for a constant force, [tex] W = \vec F \cdot \vec s [/tex] where the dot is the vector dot product using the total distance s. But here (in this problem) gravity points straight down. The component of s that is parallel to the force is delta h, the height of the stairs!]
     
    Last edited: May 25, 2010
  4. May 25, 2010 #3

    PhanthomJay

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    the units for work are newton-meters, or Joules
    That's right, this is the work done the climber against gravity, at assumed constant velocity.
    Why? Solve for delta h.

    Edit: confirming collinsmark reply.
     
  5. May 25, 2010 #4
    I'm not quite sure that I get it :(

    So when I find the force that will be my height? This is the most confusing question to me lol
     
  6. May 25, 2010 #5
    How do I find delta h?
     
  7. May 25, 2010 #6

    collinsmark

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    You had it in your original post. Earlier (in my previous post) I was simply confirming that your approach was correct. :smile:

    You had already calculated,

    [tex] W = P\Delta t [/tex]

    such that W = 2375 J. But you also worked out that

    [tex] W = mg \Delta h [/tex],

    therefore,

    [tex] \mbox{2375 J} = mg \Delta h [/tex].

    Solve for [itex] \Delta h [/itex], where [itex] \Delta h [/itex] is the height of the stairs.

    [I didn't mean to confuse you with the vector dot product stuff. All I was trying to point out was that it's not necessary to know the 3-dimensional length of the stairs. Since gravity is the only external force involved (well, besides normal forces and such) (and I'm also assuming that the student is ascending the stairs at a constant velocity [i.e. not accelerating] for simplicity), then gravity is the only force the student is "fighting." That means the height of the stairs (and not its overall length) is the only distance that matters.]
    [Edit: Oh, and I'm also ignoring air resistance and the like. :wink:]
     
    Last edited: May 25, 2010
  8. May 25, 2010 #7
    Thank you :)
    so since
    2375 J =mg(h) do I need to rearrange it so its h= ?
     
  9. May 25, 2010 #8
    m=65
    g=(9.81)
    (h)=?
    J= 2375

    I sort of it get it but still not sure.. terrible at physics
     
  10. May 25, 2010 #9
    h=(2375J)(9.81) ?
     
  11. May 25, 2010 #10

    cepheid

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    No, if you do anything to an equation, you have to do it to both sides of the equation (if you don't, then the two sides won't be equal anymore). For the equation:

    W = mgh

    ...in order to leave just h on the right hand side, you need to divide the right hand side by mg. Therefore, you need to divide the left hand side by mg as well.
     
  12. May 25, 2010 #11
    2375 J =mgh

    (65)(9.81)=637.65
    2375J/637.65
    =3.72?
     
  13. May 26, 2010 #12

    collinsmark

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    Yeh! Your 3.72 m result looks good to me (but don't forget your units). :tongue:
     
  14. May 26, 2010 #13
    Thank you so much! :) :) :)
     
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