# Work done by General Variable force

• 1MileCrash
In summary, the problem involves finding the work done by a force, given by F = F0(x/x0 - 1), in moving a particle from x = 0 to x = 2x0. The force is directed along the x-axis and x0 is an undetermined constant. The integral of the force is calculated from 0 to 0, resulting in a value of 0.
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

## 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.

Last edited:
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

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.

## What is work done by a general variable force?

Work done by a general variable force is the measurement of energy transfer from one object to another due to the application of a force that changes in magnitude and/or direction.

## How is work done by a general variable force calculated?

The calculation of work done by a general variable force involves integrating the force function with respect to the displacement of the object. The resulting value is the amount of work done.

## What is the difference between positive and negative work done by a general variable force?

Positive work is done when the applied force and the displacement of the object are in the same direction. This means that energy is transferred to the object, increasing its kinetic energy. Negative work is done when the force and displacement are in opposite directions, resulting in a decrease in the object's kinetic energy.

## How does the angle between the force and displacement affect the work done by a general variable force?

The angle between the force and displacement affects the amount of work done. When the force and displacement are perpendicular, the work done is zero because the force does not cause a change in the object's displacement. When the force and displacement are parallel, the work done is at its maximum value.

## What are some real-world examples of work done by a general variable force?

Examples of work done by a general variable force include pushing a shopping cart up a ramp, lifting weights at the gym, and throwing a ball. In each of these situations, the force applied changes in magnitude and/or direction, resulting in work being done on the object.

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