Two crates, of mass m1 = 60 kg and m2 = 130 kg, are in contact and at rest on a horizontal surface (Fig. 4-54). A 620 N force is exerted on the 60 kg crate. The coefficient of kinetic friction is 0.15.

There is no question here. Just a bunch of numbers.

(a) Calculate the acceleration of the system.

(b) Calculate the force that each crate exerts on the other
(c) Repeat with the crates reversed. fIND ACCELERATION AND CRATE FORCE.

To find the acceleration of the system and the force exerted on the 130 kg crate, we can follow these steps:

Step 1: Find the friction force acting on the system.
The friction force can be found using the equation F_friction = μ * N, where μ is the coefficient of kinetic friction and N is the normal force acting on the system. In this case, we assume the normal force is equal to the weight of the system (mg).
N = (m1 + m2) * g
F_friction = μ * N

Step 2: Find the net force acting on the system.
The net force is the force applied minus the friction force.
F_net = F_applied - F_friction

Step 3: Find the acceleration of the system.
Since the crates are in contact and at rest, the acceleration of the system can be found using Newton's second law: F_net = (m1 + m2) * a.
a = F_net / (m1 + m2)

Step 4: Find the force exerted on the 130 kg crate.
Since the crates are in contact, they have the same acceleration. Thus, the force exerted on the 130 kg crate can be found using Newton's second law: F_crate2 = m2 * a.

Now, let's calculate the values:

Step 1: Find the friction force.
N = (60 kg + 130 kg) * 9.8 m/s²
N = 1914 N (approximately)
F_friction = 0.15 * 1914 N
F_friction ≈ 287.1 N

Step 2: Find the net force.
F_applied = 620 N
F_net = 620 N - 287.1 N
F_net ≈ 332.9 N

Step 3: Find the acceleration.
a = 332.9 N / (60 kg + 130 kg)
a ≈ 1.62 m/s²

Step 4: Find the force exerted on the 130 kg crate.
F_crate2 = 130 kg * 1.62 m/s²
F_crate2 ≈ 210.6 N

Therefore, the acceleration of the system is approximately 1.62 m/s², and the force exerted on the 130 kg crate is approximately 210.6 N.