Two blocks, m1 = 4 kg and m2 = 2 kg, are in contact with each other and are on a horizontal surface for which the coefficient of kinetic friction is 0.1. A 34 N horizontal force acts on block 1, making both blocks move in the direction of block 2. 34N --> (m1)(m2) -->

Find:
a) the acceleration of the blocks
b) the force exerted by m2 on m1.

I have Ff(total) = 5.88N => Fnet = ma => a = 4.6866m/s2

I don't konw how to do b)

net force=total mass*a

34-(m1+m2)g*mu=(m1+m2)a
solve for a

force on 2:
you know a.
net force=m2*a
F-m2*g*mu=m2*a
solve for F

To find the force exerted by m2 on m1, we can use Newton's third law of motion, which states that every action has an equal and opposite reaction. This means that the force exerted by m2 on m1 is equal in magnitude but opposite in direction to the force exerted by m1 on m2.

Since there is a 34 N horizontal force acting on block 1, this force is also being exerted on block 2 due to their contact. Therefore, the force exerted by m1 on m2 is 34 N in the opposite direction.

To find the force exerted by m2 on m1, we use the equation:

\( F_{m1-m2} = - F_{m2-m1} \)

Where \( F_{m1-m2} \) is the force exerted by m1 on m2 and \( F_{m2-m1} \) is the force exerted by m2 on m1.

In this case, \( F_{m1-m2} = -34 \) N.

Therefore, the force exerted by m2 on m1 is 34 N in the same direction as the applied force.