Checking to see if I'm right or not.

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Homework Help Overview

The discussion revolves around calculating the work done by gravity on a jogger running up a hill, which is specified to be 100 m long and inclined at an angle of 12 degrees. The problem involves concepts from mechanics, particularly work and gravitational potential energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use the formula for gravitational potential energy but questions the correctness of their calculation and the expected result. Some participants discuss the integral form of work and the need to consider the component of distance in the direction of the gravitational force.

Discussion Status

Participants are exploring different interpretations of the work done by gravity, with some providing insights into the mathematical formulation of work and its dependence on the direction of force and distance. There is an acknowledgment of confusion regarding the calculations and the underlying concepts.

Contextual Notes

There is a mention of the jogger's mass and the hill's dimensions, but the original poster expresses uncertainty about the calculations and the expected outcome, indicating a potential misunderstanding of the concepts involved.

Pinkshell4u
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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 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
 

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