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.