Work done by General Variable force

AI Thread Summary
The discussion revolves around calculating the work done by a variable force defined as F = F0(x/x0 - 1) when moving a particle from x = 0 to x = 2x0. The integral for work is set up as W = ∫ from 0 to 2x0 of F0(x/x0 - 1) dx. Participants clarify that x0 is a constant and should not be treated as a limit of integration. The conclusion is that the integral evaluates to zero, indicating no work is done in this scenario. The conversation emphasizes understanding the limits of integration and the nature of the force involved.
1MileCrash
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Homework Statement



The force on a particle is directed along an x-axis and is given by F = F0(x/x0 - 1).

Find the work done by the force in moving the particle from x = 0 to x = 2x0

Homework Equations





The Attempt at a Solution



It looks like the force is a recurrence relation or something...

W = \int^{2x_{0}}_{0} F_{0}(x/x_{0}-1) dx

I don't really understand.. unless I just take x0 to be the lower limit of integration, 0, and then 2x0 is just also 0, leaving me integrating from 0 to 0... which is just 0, but that seems to render this an absurdly stupid question.
 
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F0 and x0 are undetermined constant.

ehild
 
So x0 is could not be described as lower limit of integration? I see it as integrating from original x position to two times original x position, which in this case is 0 to 0, which means the integral is 0?

How do you view it?
 
1MileCrash said:
So x0 is could not be described as lower limit of integration? I see it as integrating from original x position to two times original x position, which in this case is 0 to 0, which means the integral is 0?

How do you view it?

X0 is just a number and the upper limit of integration is 2X0, the lover limit is zero.

Yes, the integral is 0. The antiderivative is F0(x2/(2x0)-x), which returns 0 between the limits.

ehild
 
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