Please help me with this conceptual Work problem?

Question

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 shown in the figure.

http://session.masteringphysics.com/problemAsset/1011078/29/MWE_cw_5_a.jpg

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,g ,L ,mu1 , and mu2.

I would really appreciate help with this one guys, I keep getting it wrong :( Thanks in advance!

I forgot to add this second part of the problem. I don't know why it didn't copy over.

What is the total work done by the external force in pulling the board from region 1 to region 2? (Again, assume that the board moves at constant velocity.)

Answer 1

ρ = M/L
s = distance into regioin 2 from boundary
W2 = ρgs
W1 = ρg(L - s)
F = μ1W1 + μ2W2
F = μ1ρg(L - s) + μ2ρgs
Fds = (Mg/L)(μ1(L - s) + μ2s)ds
W = (Mg/L)(μ1(L^2 - (1/2)L^2) + (1/2)μ2L^2)
W = (MgL)(μ1 - (1/2)μ1 + (1/2)μ2)
W = (MgL)(μ1 + μ2)/2

Answer 2

Work = force times distance,
Distance = L
Force is the average of the two friction forces:
Av Force = M*g*(mu1+mu2)/2
so the work done pulling the board from one side to the other is:
W=L*M*g*(mu1+mu2)/2

The "net work done by friction" i don't understand. I guess one would argue that friction trasformed the above amount of energy into heat, so it is the same.

Answer 3

g is negative, and the net work done by the external force is the same, but positive because net work should equal 0.

Answer 4

W = -(MgL)(μ1 + μ2)/2

Answer 5

The answer is [-(MgL)(u_1+u_2)]/2