Integrating F.dx: Constant μ, m & g Affect Force?

[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?

 

[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).

 

[Hesperides]

Great! Thanks Drew

 

[Naty1]

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

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

 

[PJC01]

I can confirm that your understanding is correct. In this scenario, the friction force is not affected by integration because the variables μ, m, and g are all considered to be constants. When integrating x^0, the result is simply x, which is then multiplied by the constant μ, m, and g. Since these are all constants, the friction force remains unchanged and does not require further integration. This is because integration is used to find the total change in a variable over a given range, but in this case, the variables are already constants and do not change over the given range. Therefore, the integration process does not affect the friction force.