# Integrating F.dx

1. Oct 20, 2012

### Hesperides

I'm sure this is an easy question to answer - per the attached image, can someone please explain why the friction force has not required/been affected by integration?

My calculus is very rusty but the only reason I can think of is that μ, m and g are all constants (for the purposes of the question) and so x^1 drops in as a result of integrating x^0 - is that correct?

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2. Oct 20, 2012

### DrewD

Yup. The force of friction is the same at every point along the path for the reasons that you stated. Since integration is just adding up all of those amounts of friction at each point, it is the same as multiplying it by $x$. In reality, as you can probably guess, the force of friction would not actually be totally constant, but the fluctuations are very small (unless you allow the block to stop at some point at which point you would need to worry about static friction).

3. Oct 20, 2012

### Hesperides

Great! Thanks Drew

4. Oct 21, 2012

### Naty1

It's a line integral over a distance x2 - x1.....

so the constants may be moved outside the integral leaving just the displacement.