Checking to see if I'm right or not.

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Checking to see if I'm right...or not.

Homework Statement


A hill is 100 m long and makes an angle of 12 degrees with the horizontal. As a 50-kg jogger runs up the hill, how much work does gravity do on the jogger?


Homework Equations



Do I use PEg=mgDelta Y ?


The Attempt at a Solution



PEg = (50 kg) (9.8 m/s2) (100m) which equals 49,000 which is wrong. I know the answer is -10,000 J.

How and why? So confused...
 
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work is

[tex]\int\vec{F}\bullet dr[/tex]

to get rid of the dot product, you must take the parallel component of the distance that it is in the same direction of the force (gravity..which is down)

so you get :

-F*d*sin([tex]\theta[/tex]) = -mg*d*sin([tex]\theta[/tex]) = -(50)(9.8)*100*sin(12)
 


Oh, I think I get it. Thanks a lot.
 


no problem, remember that when you have the expression for work as:

F * D

that D is the change in distance, so say you climb a mountain than you climb back down than the work overall is zero since the change in position is zero