"A uniform board of length L and mass M lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is (mu1), and in region 2, the coefficient is (mu2). The positive direction is from the region with mu1 to the region with mu2. Find the net work W done by friction in pulling the board directly from region 1 to region 2. Assume that the board moves at constant velocity. Express the net work in terms of M, L, g, mu1 and mu2"(adsbygoogle = window.adsbygoogle || []).push({});

In my force formula, x is equal to the amount of the board in region 2 (with mu2).

F = mu1(mg(1-(x/L))) + mu2(mg(x/L)

I know I'm supposed to integrate this with respect to L (or am I). Unfortunately, I suck at calculus. This is what I got, sorta.

((1/2)((mu1*g)/L)L^2) + ((1/2)((mu2*g)/L)L^2

Again, I suck at calc. Any thoughts?

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# Gragging a board across surfaces with different frictions

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