Work done by General Variable force

[1MileCrash]

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...

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

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.

 

[ehild]

F₀ and x₀ are undetermined constant.

ehild

 

[1MileCrash]

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?

 

[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?

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

 

[asteriatic]

I would approach this problem by first clarifying any ambiguities in the given information. It is unclear if x0 is a constant or a variable in this situation. Additionally, the given force equation does not specify any units, which could impact the calculation of work.

Assuming x0 is a constant and the force is in Newtons, the work done by this force in moving the particle from x = 0 to x = 2x0 can be calculated as follows:

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

= F0 \int^{2x_{0}}_{0} (x/x_{0}-1) dx

= F0 \left[\frac{x^2}{2x_{0}} - x\right]^{2x_{0}}_{0}

= F0 \left[\frac{4x_{0}^2}{2x_{0}} - 2x_{0}\right]

= 2F0x_{0} - 2F0x_{0}

= 0

Therefore, the work done by this force in moving the particle from x = 0 to x = 2x0 is 0. This result may seem counterintuitive, but it makes sense because the force is always perpendicular to the displacement of the particle, resulting in no work being done. However, if the force equation or the limits of integration were different, the work done could have a non-zero value. It is important to carefully consider all factors and variables in a problem before attempting to solve it.