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 μ₁, and in region 2, the coefficient is μ₂. The positive direction is shown in the figure.

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.)Express your answer in terms of M, g, L, μ_rm1, and μ_rm2.

Answer 1

work done=mg*μ1*L+mg*μ2*L

Answer 2

suppose length x lies in region 1 and length y in region 2 where x+y=L.
initially x=a.
i assume the board is pulled normal to the boundary.

weight of board in region 1 is M(x/L)g and in region 2 is M(y/L)g.
the frictional resistance forces are M(x/L)gu1 and M(y/L)gu2.

to get the board out of region 1 it must be moved a distance a.

work done = force x distance.
but the force is changing so the work done will be an integral.
work done W = integral (F dx)
where total force F depends on distance x, and the integral is from x=a to 0.

F = M(x/L)gu1 + M(y/L)gu2
= Mg ( (x/L)u1 + (1-x/L)u2)
= Mg ( u2 + (u1-u2)(x/L) )

integral (F dx) = Mg ( xu2 + (u1-u2)(x^2/2L) )

when x=a,
integral = Mg ( au2 + (u1-u2)(a^2/2L) )
when x=0,
integral = 0.

so W = Mga ( u2 + (u1-u2)(a/2L) )

if eg the board is initially half in region 1 (a=L/2) then
W = (MgL/2) ( u2 + (u1-u2)(1/4) )
= (MgL/8) ( 4u2 + (u1-u2) )
= (MgL/8)( 3u2 + u1 ).

instead of integrating, since the force varies linearly with distance, you could use
work done = average force x distance
distance = a
initial force (x=a) is Mg ( u2 + (u1-u2)(a/L) )
final force (x=0) is Mg (u2)
average force during motion
= (1/2)Mg ( 2Lu2 + (u1-u2)(a/L) )
= Mg ( Lu2 + (u1-u2)(a/2L)
work done
W = Mg ( Lu2 + (u1-u2)(a/2L) )a
=Mga ( Lu2 + (u1-u2)(a/2L)
as above.