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