(II) Two crates, of mass 75 kg and 110 kg, are in contact and at rest on a horizontal surface. A 620-N force is exerted on the 75-kg crate. If the coefficient of kinetic friction is 0.15, calculate (a) the acceleration of the system, and (b) the force that each crate exerts on the other.
(c) Repeat with the crates reversed.
To solve this problem, we can use Newton's second law of motion and apply it to each crate separately.
(a) To find the acceleration of the system, we first need to determine the net force acting on the system. The net force is the difference between the applied force and the frictional force.
The frictional force can be calculated using the equation: Frictional force = coefficient of kinetic friction * normal force.
For the 75 kg crate:
Frictional force1 = 0.15 * (mass1 * gravitational acceleration)
= 0.15 * (75 kg * 9.8 m/s^2)
= 110.25 N
For the 110 kg crate, the normal force is equal to the weight of both crates:
Frictional force2 = 0.15 * (mass2 * gravitational acceleration)
= 0.15 * (110 kg * 9.8 m/s^2)
= 161.7 N
Now, let's calculate the net force on the system:
Net force = applied force - frictional force1 - frictional force2
= 620 N - 110.25 N - 161.7 N
= 348.05 N
Since the acceleration is the net force divided by the total mass of the system:
Acceleration = Net force / (mass1 + mass2)
= 348.05 N / (75 kg + 110 kg)
= 348.05 N / 185 kg
= 1.88 m/s^2
Therefore, the acceleration of the system is 1.88 m/s^2.
(b) To find the force that each crate exerts on the other, we can apply Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
For the 75 kg crate:
Force on crate1 = mass1 * acceleration
= 75 kg * 1.88 m/s^2
= 141 N
For the 110 kg crate:
Force on crate2 = mass2 * acceleration
= 110 kg * 1.88 m/s^2
= 206.8 N
Therefore, the force that each crate exerts on the other is 141 N and 206.8 N.
(c) When the crates are reversed, the applied force will now act on the 110 kg crate. The mass and coefficient of friction remain the same, so we can use the same calculations as above. The only difference is that we need to change the applied force from 620 N to be acting on the 110 kg crate.
(a) The acceleration of the system remains the same, which is 1.88 m/s^2.
(b) The force that each crate exerts on the other remains the same, which is 141 N and 206.8 N.