Force&Friction- Two blocks stacked on top

[physStudent_3]

Hello, I have worked out this physics problem. I then checked my answer, it was correct. However, I have a question about Part B which I will address once all of my information is given;

Homework Statement

A slab of mass m2 = 40kg, rests on a frictionless floor, and a block of mass m1 = 10kg rests on top of the slab. Between the block and the slab the coefficient of static friction is 0.60 and the coefficient of kinetic friction is 0.40. The block is pulled by a horizontal force F of magnitude 100N. What are the resulting accelerations of (a) the block and (b) the slab?

2. Relevant data
m1 = mass of block
m2 = mass of slab
f1 = force of friction between block and slab
a1 = acceleration of block
a2 = acceleration of slab

The Attempt at a Solution

A) a1 = [ F - (u)(m1)(g) ] / (m1)

a1 = [ 100 - (0.4)(10)(9.8) ] / (10)

a1 = 6.08 m/s^2

B) a2 = f1 / m2

a2 = 39.2 / 40

a2 = 0.98 m/s^2

My question is; Since the block (m1) is resting on top of the slab (m2) then shouldn't the two be added together in the denominator of the equation in part B?

Is my solution manual wrong?

Thank you for your insight.

 

[ashishsinghal]

Is the answer given under "attempt at solution" solution manual's or yours?
That part seems correct.

 

[ashishsinghal]

My question is; Since the block (m1) is resting on top of the slab (m2) then shouldn't the two be added together in the denominator of the equation in part B?

Is my solution manual wrong?

no, that is wrong:
firstly because in part B the force is friction force which will become internal force and hence cannot be used.
secondly, you cannot club the bodies because they are moving with respect to each other and hence on clubbing it would not remain a translational motion