Calculating Joe's Climb Efficiency

[lcp1992]

Homework Statement

Joe ascends a small hill from an elevation of 250m to an elevation of 490m. In doing so he expends 2.8x10^5J. If Joe's mass is 75kg what is the efficiency of his climb?

Homework Equations

W= change in E= change in Ek + change in Ep
Ep= mgh
Ek= 0.5mv^2
%efficiency=useful energy output/total energy input x 100%

The Attempt at a Solution

should i use the work energy theorem to solve the problem?

 

[Delphi51]

Welcome to PF!
I wouldn't use the theorem since there is no kinetic energy involved. The "useful energy" is just mgh.

 

[lcp1992]

Iused the mgh formula, but I couldn't get the answer which is 0.63

 

[dacruick]

If the climber has gained the energy 'mgh' from climbing the hill, and the energy exerted is 2.8x10^5J, then his efficiency would be what?

 

[lcp1992]

0.63 is the efficiency

 

[Delphi51]

So you got it? It certainly works out to .63 for me. If it didn't for you, show your work.

 

[lcp1992]

I got it but I am still confused so the work is total energy input and the energy is the useful energy , why?

 

[cepheid]

I got it but I am still confused so the work is total energy input and the energy is the useful energy , why?

You only need to do mgh amount of work to climb a hill up to height h. But human beings (like most machines) are inefficient, which means that they will actually use MORE energy than this during the trip. The excess energy is wasted (as heat or whatever). That's why the input energy is larger than the output (useful) energy which actually went into doing the work necessary to get up the hill.

 

[lcp1992]

thanks!