Final answer:
The acceleration of the system with two crates having masses 75 kg and 110 kg when a force of 730 N is applied on one of them, considering a coefficient of kinetic friction of 0.15, is approximately 2.47 m/s².
Step-by-step explanation:
The total mass of the system is the sum of the masses of the two crates, which is 185 kg.
The net force, taking into account friction, is the applied force of 730 N minus the force of kinetic friction. The force of kinetic friction is determined by multiplying the normal force (the total weight of the system, or mass times gravity) by the coefficient of kinetic friction, which is 0.15. The normal force in this case is 185 kg * 9.8 m/s² = 1813 N, so the force of friction is 0.15 * 1813 = 272 N.
Now, subtract this frictional force from the applied force to determine the net force: 730 N - 272 N = 458 N. Dividing this net force by the total mass gives the acceleration of the system. Therefore, 458 N / 185 kg = 2.47 m/s². So, the acceleration of the system is approximately 2.47 m/s².
Learn more about Newton's Laws of Motion