What is the work done by a hiker who is climbing up a hill

AI Thread Summary
The discussion revolves around calculating the work done by a hiker climbing a hill, defined by the function f(x). The work is expressed as W = ∫[a,b] F dx, where F is the force acting on the hiker. Participants clarify that if the force is constant, work can be simplified to W = F * Δx, with Δx derived from the difference in elevation, f(b) - f(a). There is some confusion about whether to integrate over the force with respect to the hill's equation or the elevation change. Ultimately, the consensus is that the work done primarily depends on the change in elevation, leading to the formula W = mg * Δx.
Weave
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Homework Statement


A hiker of mass(m) climbs up a hill( the equation of the hill is f(x), what is the work done on the interval [a,b]

Homework Equations


W=\int_{a}^{b}F dx

The Attempt at a Solution


So for this problem I wanted to make sure. We would do:
Assuming Force is constant, we would have W=Fdelta x
delta x can be obtained from f(b)-f(a).
If the force is not constant we do
W=\int_{f(a)}^{f(b)}F dx
Right?
 
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Is the elevation a function of x? Neglecting friction, work would just be mgdeltax
 
I am not sure, my Prof. said that the equation is just the equation of the hill.
 
Shoudn't it be W=\int_{a}^{b}F dx
instead of W=\int_{f(a)}^{f(b)}F dx?
Because work is just area under the force-displacement graph.

I think, for x, x is the horizontal distance and f(x) is the vertical distance.
 
rootX said:
Shoudn't it be W=\int_{a}^{b}F dx
instead of W=\int_{f(a)}^{f(b)}F dx?
Because work is just area under the force-displacement graph.

I think, for x, x is the horizontal distance and f(x) is the vertical distance.
So how about the equation of the hill- f(x)?
 
so...
 
Weave said:
So how about the equation of the hill- f(x)?

:shy:
I dunn know, I was just guesing
:cry:
 
My guess is that the force is constant. I would agree with turdferguson.
Work done depends only on the change in elevation.
So W=mg*delta x
 
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