How Many Beers Does It Take to Reimburse Energy Expenditure?

[nns91]

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=$$\Delta$$K

The Attempt at a Solution

So E= F.$$\Delta$$x =mgx= 62*9.81*2

For 676 times E=672*62*9.81*2$$\approx$$ 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 ?

 

[fluidistic]

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.

 

[mburt3]

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.

 

[nns91]

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 ?

 

[mburt3]

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.

 

[nns91]

so will normal force equal force of gravity + Fsin30 ?

 

[nns91]

Actually, subtract I meant. Anyway, I got it. THanks guys.