Three blocks on a frictionless horizontal surface are in contact with each other. A force F is applied to block A and the blocks start to slide to the right of their initial position. Determine the acceleration of the system, if m_A = 12 kg, m_B = 38 kg, m_C = 4700 g, and F = 198 N.

Draw a free-body diagram for each block.

a_{net} =

Acceleration of the system is 3.6 m/s^2

To determine the acceleration of the system, we need to analyze the forces acting on each block using the free-body diagram. Let's first draw a free-body diagram for each block.

For block A:
- The force applied, F, is acting to the right.
- The weight of block A, mg_A, is acting downward.
- The normal force, N_A, from block B is acting upward and perpendicular to the surface.

For block B:
- The weight of block B, mg_B, is acting downward.
- The normal force, N_B, from block A is acting upward.
- The normal force, N_C, from block C is acting upward and perpendicular to the surface.

For block C:
- The weight of block C, mg_C, is acting downward.
- The normal force, N_C, from block B is acting upward.

Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Let's calculate the net force for each block.

For block A:
- The net force is given by F_net_A = F - N_A.

For block B:
- The net force is given by F_net_B = N_A - N_B.

For block C:
- The net force is given by F_net_C = N_B - N_C.

Since there is no friction between the blocks, the normal forces are equal in magnitude to the weight of the blocks. Therefore:

N_A = mg_A,
N_B = mg_B,
N_C = mg_C.

Substituting these values into the equations for net force, we get:

F_net_A = F - mg_A,
F_net_B = mg_A - mg_B,
F_net_C = mg_B - mg_C.

Now, we can calculate the net forces:

F_net_A = 198 N - (12 kg)(9.8 m/s^2) = 198 N - 117.6 N = 80.4 N,
F_net_B = (12 kg)(9.8 m/s^2) - (38 kg)(9.8 m/s^2) = 117.6 N - 372.4 N = -254.8 N,
F_net_C = (38 kg)(9.8 m/s^2) - (4.7 kg)(9.8 m/s^2) = 372.4 N - 46.06 N = 326.34 N.

Since the blocks are in contact and connected, they move together as a system. Therefore, the acceleration of the system is the same for all blocks. We can calculate the acceleration using the equation:

F_net_total = (m_A + m_B + m_C) * a,

where a is the acceleration of the system. Substituting the values, we get:

80.4 N + (-254.8 N) + 326.34 N = (12 kg + 38 kg + 4.7 kg) * a,

151.94 N = 54.7 kg * a.

Dividing both sides by 54.7 kg, we get:

a = 151.94 N / 54.7 kg ≈ 2.77 m/s^2.

Therefore, the acceleration of the system is approximately 2.77 m/s^2 to the right.