Two crates connected by a rope of negligible mass lie on a horizontal surface. Crate has mass and crate has mass . The coefficient of kinetic friction between each crate and the surface is Micro k. The crates are pulled to the right at a constant velocity of 3.2 by a horizontal force . Part A) In terms of Ma, Mb, and Micro k, calculate the magnitude of the force .
Part B) In terms of Ma, Mb, and Micro k, calculate the tension in the rope connecting the blocks.

a)    F = μ g (M_a + M_b),  b)   T = μ g M_b

Explanation:

a) as the box goes at constant speed, the relation is zero and we can use the translational equilibrium relation

              ∑ F = 0

Y axis

              N₁ - W₁ = 0

              N₂-W₂ = 0

              N₁ = W₁

              N₂ = W₂

X axis

              F -fr₁ -fr₂ = 0

              F = fr₁ + fr₂

the friction force has the expression

              fr = μ N

indicates that the friction coefficient is the same for the two boxes

               fr₁ = μ M_a g

               Fr₂ = μ M_b g

we substitute

                F = μ g (M_a + M_b)

b) let's apply the equilibrium for box A

                 F -fr₁ - T = 0

                 T = F -fr1

boc B

                T - fr2 =0

                T = fr2

our system equation

                 T = F -fr1

                 T = fr2

solved

                2 T = F - fr₁ + fr₂

                2T = [μ g (M_a + M_b) - μ g M_a + μ g M_b]

                2T =  2 μ g M_b

                  T = μ g M_b

the answer is b i think