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Conservation of energy

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A man lifted a full keg of beer (mas 62kg) to a height of about 2m 676 times in 6 hours. Assuming that work was done only as the keg was going up, estimate how many such kegs of beer he would have to drink to reimburse his energy expenditure. (1 L of beer is approximately 1kg and provides about 1.5MJ of energy; in calculation neglect the mass of the empty keg)

    2. Relevant equations


    3. The attempt at a solution

    So E= F.[tex]\Delta[/tex]x =mgx= 62*9.81*2

    For 676 times E=672*62*9.81*2[tex]\approx[/tex] 822313 J.

    So the energy he needs to take in is 822313 J which is smaller than 1.5MJ so it means he will need less than 1 L of beer ??

    Am I wrong ?
  2. jcsd
  3. Dec 2, 2008 #2


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    Gold Member

    I think you made it right even if you wrote "672" instead of 676 in
  4. Dec 2, 2008 #3
    Amusing problem by the way. I think that you did it right, I got that it was 0.5482 L or 0.008842 kegs of beer.
  5. Dec 2, 2008 #4
    My bad. So my approach is right ??

    One more problem:

    So I am given a 8kg sled is initially at rest on a horizontal road. The coefficient of kinetic friction between the sled an the road is 0.4. The sled is pulled a distanced of 3m by a force of 40 N applied to the sled at an angle of 30 degree.

    They ask for the energy dissipated by friction

    So I have delta E thermal = F* delta x= 0.4*8*9.81*3

    However I got the wrong answer. Is my approach right ?
  6. Dec 2, 2008 #5
    The normal force is not equal to the force of gravity because there is another force acting in the vertical direction---the force being applied to the sled at the 30 degree angle.
  7. Dec 2, 2008 #6
    so will normal force equal force of gravity + Fsin30 ?
  8. Dec 2, 2008 #7
    Actually, subtract I meant. Anyway, I got it. THanks guys.
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