A 6.00-kg block is in contact with a 4.00-kg block on a horizontal frictionless surface as shown in the figure.The 6.00-kg block is being pushed by a horizontal 20.0-N force as shown. What is the magnitude of the force that the 6.00-kg block exerts on the 4.00-kg block?
A) 6.00 N
B) 20.0 N
C) 8.00 N
D) 4.00 N
E) 10.0 N

Question

A 6.00-kg block is in contact with a 4.00-kg block on a horizontal frictionless surface as shown in the figure.The 6.00-kg block is being pushed by a horizontal 20.0-N force as shown. What is the magnitude of the force that the 6.00-kg block exerts on the 4.00-kg block? 
A) 6.00 N 
B) 20.0 N 
C) 8.00 N 
D) 4.00 N 
E) 10.0 N

Quizplus · Accepted Answer

Since the surface is frictionless, the two blocks will move together with the same acceleration. Therefore, we assume a common acceleration, a. 
From F = ma, we can calculate the acceleration as:

a = F / (m1 + m2) = 20.0 N / (6.00 kg + 4.00 kg) = 2.00 m/s^2

The force that the 6.00-kg block exerts on the 4.00-kg block is equal in magnitude but opposite in direction to the force applied by the 4.00-kg block on the 6.00-kg block due to Newton's 3rd Law. 
Thus, we can calculate this force as:

F12 = m2 * a = 4.00 kg * 2.00 m/s^2 = 8.00 N

Therefore, the answer is (C) 8.00 N.

A 600-Kg Block Is in Contact with a 4