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)
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 ?