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Find efficiency

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Joe ascends a small hill from an elevation of 250m to an elevation of 490m. In doing so he expends 2.8x10^5J. If Joe's mass is 75kg what is the efficiency of his climb?

    2. Relevant equations
    W= change in E= change in Ek + change in Ep
    Ep= mgh
    Ek= 0.5mv^2
    %efficiency=useful energy output/total energy input x 100%

    3. The attempt at a solution
    should i use the work energy theorem to solve the problem?
  2. jcsd
  3. Nov 24, 2011 #2


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    Homework Helper

    Welcome to PF!
    I wouldn't use the theorem since there is no kinetic energy involved. The "useful energy" is just mgh.
  4. Nov 24, 2011 #3
    Iused the mgh formula, but I couldn't get the answer which is 0.63
  5. Nov 24, 2011 #4
    If the climber has gained the energy 'mgh' from climbing the hill, and the energy exerted is 2.8x10^5J, then his efficiency would be what?
  6. Nov 24, 2011 #5
    0.63 is the efficiency
  7. Nov 24, 2011 #6


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    Homework Helper

    So you got it? It certainly works out to .63 for me. If it didn't for you, show your work.
  8. Nov 24, 2011 #7
    I got it but I am still confused so the work is total energy input and the energy is the useful energy , why?
  9. Nov 24, 2011 #8


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    Gold Member

    You only need to do mgh amount of work to climb a hill up to height h. But human beings (like most machines) are inefficient, which means that they will actually use MORE energy than this during the trip. The excess energy is wasted (as heat or whatever). That's why the input energy is larger than the output (useful) energy which actually went into doing the work necessary to get up the hill.
  10. Nov 24, 2011 #9
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