## Confusing Force function

1. The problem statement, all variables and given/known data
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 $$x = 2x_0$$

2. Relevant equations
F=ma, W=Fd, etc.

3. 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.
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Hi DrummingAtom!

(try using the X2 tag just above the Reply box )
 Quote by DrummingAtom 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! )
 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.

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## Confusing Force function

 Quote by 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.
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.