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Homework Help: Work and enegry theoretical problem

  1. Mar 19, 2006 #1
    Ok so my problem was: A 80 kg hiker starts at an elevation of 1600 m and climbs to the top of a 3600 m peak.

    (a) What is the hiker's change in potential energy?

    (b) What is the minimum work required of the hiker?

    (c) Can the actual work done be more than this? YES!!

    so i figured out a) Change in PE = mg(y2-y1)
    =80kg(9.8 m/s^2)(3600m-1600m)
    = 1.56 x 10^6 Joules
    B) Part b is the same answer cause with your PE you can perform minimum work

    But for C i dont understand how the actual work be more than delta PE?
  2. jcsd
  3. Mar 19, 2006 #2
    Are you considering friction?
  4. Mar 19, 2006 #3
    Ok ..if we consider friction t the opposite force ( energy used to climb up the peak) has to be more than 1.5 e6 because it has to overcome the force of friction to kepp climbing up. is that right?
  5. Mar 19, 2006 #4
    Yes. That's what I was thinking.
  6. Mar 19, 2006 #5
    thanx so much for your help
  7. Mar 19, 2006 #6
    It was a pleasure, but you should ask a second and better opinion.
  8. Mar 19, 2006 #7
    im going to ask my professor in class thanx anyway
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