Calculating Work Done by a Confusing Force Function | F=ma, W=Fd

[DrummingAtom]

Homework Statement

The force on a particle is directed along an x-axis and given by $$F = F_0(\frac {x}{x_0} -1)$$. Find the work done by the force in moving the particle from x = 0 to $$x = 2x_0$$

Homework Equations

F=ma, W=Fd, etc.

The Attempt at a Solution

I don't even know how to interpret that function. Does the $$x_0$$ mean the initial position? Does $$F_0$$ mean the initial force? I'm so confused. Any help would be appreciated.

 

[tiny-tim]

Hi DrummingAtom!

(try using the X₂ tag just above the Reply box )

I don't even know how to interpret that function. Does the $$x_0$$ mean the initial position? Does $$F_0$$ mean the initial force?

That's right …

a "0" subscript always means a constant (usually the value at t = 0).

(oh … except in relativity, where x₀ means time! )

 

[DrummingAtom]

I'm still confused on this one. So, if _{x0 F0} are constants then how would the graph of this function look? Because they want you to graph F(x) before integrating. I mean what do you pick for your constant in a situation like this? I know it's going to be a linear function.

 

[collinsmark]

I'm still confused on this one. So, if _{x0 F0} are constants then how would the graph of this function look? Because they want you to graph F(x) before integrating. I mean what do you pick for your constant in a situation like this? I know it's going to be a linear function.

It doesn't really matter, as long as x₀ is not 0 (otherwise you'll have a divide by zero problem). But if you want to make your life easier, put it on the positive x-axis somewhere. I suggest putting it at x = 1. That way you'll integrate from 0 to 2. But don't label you x-axis with '1' and '2'; rather label you x-axis to go from

0...x₀...2x₀...3x₀...

Now when you consider your graph's labels, you are integrating from 0 to 2x₀, as the problem specifies!

The y-axis is F. So where does F₀ fit into your graph? I'll let you do that.

 

[rdl]

As a scientist, it is important to approach problems with a clear understanding of the variables and equations involved. In this case, the given force function can be interpreted as follows: x_0 represents the initial position of the particle, and F_0 represents the initial force acting on the particle. The force on the particle is then given by the difference between the current position (x) and the initial position (x_0) multiplied by the initial force (F_0).

To calculate the work done by this force, we can use the equation W=Fd, where F is the force and d is the displacement. In this case, the force function is given in terms of position (x), so we need to find a way to express the displacement in terms of x.

We know that the particle is moving from x=0 to x=2x_0, so the displacement can be written as d=x-0=x. Substituting this into the work equation, we get W=Fx.

Now, we can use the given force function to express F in terms of x. Plugging in F=F_0(\frac {x}{x_0} -1), we get W=F_0(\frac {x}{x_0} -1)x.

To solve for the work, we need to integrate this expression over the interval x=0 to x=2x_0. This will give us the total work done by the force in moving the particle from its initial position to its final position.

In conclusion, while the given force function may seem confusing at first, by carefully interpreting the variables and using the appropriate equations, we can calculate the work done by the force and gain a better understanding of its behavior.