Blocks with masses of 3.0 kg, 4.0 kg, and 5.0 kg are lined up in a row on a frictionless table. All three are pushed forward by a 13 N force applied to the 3.0 kg block. How much force does the 4.0 kg block exert on the 5.0 kg block?

5.42 N

Explanation:

Let the contact force between 3 kg anf 4 kg is f1 and between 4 kg and 5 kg is f2.

F = 13 N

Let a be the acceleration in the system.

Use the free body diagrams and apply Newton's second law of motion.

F - f1 = 3 x a ... (1)

f1 - f2 = 4 x a ... (2)

f2 = 5 x a ... (3)

By adding all these equation, we get

a = 13 / 12 m / s^2

Substitute the value of a in equation (3), we get

f2 = 5 x 13 / 12 = 5.42 N

Fc = 5.41 N

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

Newton's second law for the set of the three blocks

F = 13 N

m=3.0 kg + 4.0 kg+5.0 kg = 12 kg

F = m*a

13 = 12*a

a = 13 / 12

a = 1.083 m/s² : acceleration of the set of the three blocks

Newton's second law for the 5.0 kg block

m = 5.0 kg

a = 1.083 m/s²

Fc: Contact force of the 4 kg block on the 5 kg block

Fc = 5.0 kg * 1.083 m/s²

Fc = 5.41 N