Two boxes on a horizontal surface, as shown in the figure above, are moving together under the influence of a horizontal force, F. The coefficient of kinetic friction between the boxes and the surface is 0.40. Use g = 10 m/s2. In case (a), a horizontal force F directed right is applied to the larger box, which has a mass of 8.00 kg. In case (b), the horizontal force F is instead applied to the smaller box, which has a mass of 4.00 kg. The horizontal force F has a magnitude of 110. N.

(a) Let's say that the boxes are moving to the right, in the direction of the force F. What is the magnitude of the force exerted by the larger box on the smaller box in:

(i) case (a)
(N)
(ii) case (b)
(N)

(b) Now, consider the case in which the boxes are initially moving to the left, opposite to the direction of the force F.

(i) For case (a), what is the magnitude of the acceleration of the system?

(N)
(ii) For case (a), what is the magnitude of the force applied by the larger box on the smaller box?
(N )

To find the force exerted by one box on the other, we need to consider the forces acting on the system.

(a) (i) In case (a), since the boxes are moving to the right, the frictional force on the larger box will be opposing the motion. The frictional force can be calculated using the formula:

Frictional force = coefficient of kinetic friction * normal force

The normal force is the force exerted by the surface on the box, which is equal to the weight of the box (mass * gravity). Therefore, the frictional force on the larger box is given by:

Frictional force = (coefficient of kinetic friction) * (mass of larger box) * g

Plugging in the given values, we get:

Frictional force = 0.40 * 8.00 kg * 10 m/s^2 = 32.00 N

The force exerted by the larger box on the smaller box will be equal in magnitude but opposite in direction to the frictional force. Therefore, the magnitude of the force exerted by the larger box on the smaller box in case (a) is:

Force exerted by larger box on smaller box = 32.00 N

(a) (ii) In case (b), the force is applied to the smaller box. So, the magnitude of the force exerted by the larger box on the smaller box will be zero, as there is no direct contact force between them in this case.

(b) (i) In this case, since the boxes are initially moving to the left, opposite to the direction of the force F, the net force acting on the system will be the difference between the force F and the frictional force.

Net force = Force F - Frictional force

The acceleration of the system can be calculated using Newton's second law:

Net force = (mass of larger box + mass of smaller box) * acceleration

Equating the net force equations:

Force F - Frictional force = (mass of larger box + mass of smaller box) * acceleration

Plugging in the given values:

110 N - 32 N = (8.00 kg + 4.00 kg) * acceleration

78 N = 12.00 kg * acceleration

Therefore, the magnitude of the acceleration of the system in case (a) is:

Acceleration = 78 N / 12.00 kg = 6.50 m/s^2

(b) (ii) In this case, we need to calculate the force applied by the larger box on the smaller box. The force applied by the larger box is equal to the frictional force, as there are no other external forces acting on the system.

Therefore, the magnitude of the force applied by the larger box on the smaller box in case (a) is:

Force applied by larger box on smaller box = Frictional force = 32.00 N