Final answer:
The larger 110 kg crate exerts a backward force of 272.25 N on the 75 kg crate when a 730 N force is applied to it due to the acceleration caused by the net force after overcoming kinetic friction.
Step-by-step explanation:
To find the force that the 75 kg and 110 kg crates exert on each other when a 730 N force is exerted on the 75 kg crate, we must first calculate the frictional force affecting them.
The total mass of the system is 75 kg + 110 kg = 185 kg.
The normal force for the system is therefore the combined weight of both crates, i.e., (185 kg)(9.8 m/s2) = 1813 N.
Using the coefficient of kinetic friction, μk = 0.15, the total kinetic frictional force opposing the motion is μk N = 0.15 * 1813 N = 271.95 N.
Since the crates are in contact and move together as a single system, the 730 N force must overcome the frictional force for both crates.
Therefore, the net force accelerating the system is 730 N - 271.95 N = 458.05 N.
To find the acceleration, we use Newton's second law: Fnet = ma, meaning a = Fnet / m = 458.05 N / 185 kg = 2.475 m/s2.
The force that each crate exerts on the other is due to this acceleration.
The larger crate (110 kg) will exert a backward (to the 75 kg crate) force equal to its mass times the acceleration, which is (110 kg)(2.475 m/s2) = 272.25 N.