Conservation of energy

  • Thread starter nns91
  • Start date
  • #1
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Homework Statement



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)

Homework Equations



W=[tex]\Delta[/tex]K

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 ?
 

Answers and Replies

  • #2
fluidistic
Gold Member
3,790
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I think you made it right even if you wrote "672" instead of 676 in
For 676 times E=672*62*9.81*2LaTeX Code: \\approx 822313 J.
 
  • #3
23
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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.
 
  • #4
301
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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 ?
 
  • #5
23
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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.
 
  • #6
301
1
so will normal force equal force of gravity + Fsin30 ?
 
  • #7
301
1
Actually, subtract I meant. Anyway, I got it. THanks guys.
 

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