Checking to see if I'm right or not.

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AI Thread Summary
The discussion focuses on calculating the work done by gravity on a jogger running up a hill. The jogger's mass is 50 kg, and the hill is 100 m long with a 12-degree incline. The initial attempt to calculate gravitational potential energy (PEg) using the formula PEg = mgΔY was incorrect, leading to confusion about the correct approach. The correct method involves using the formula for work done by gravity, incorporating the angle of the incline, resulting in a calculation of -10,000 J. The discussion emphasizes understanding the relationship between force, distance, and the direction of movement in work calculations.
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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

\int\vec{F}\bullet dr

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(\theta) = -mg*d*sin(\theta) = -(50)(9.8)*100*sin(12)
 


Oh, I think I get it. Thanks alot.
 


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
 
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