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 blocks will move together with the same acceleration. Using Newton's second law, we can set up the following equation:

(6.00 kg + 4.00 kg) a = 20.0 N

where a is the acceleration of the blocks.

Solving for a, we get:

a = 2.00 m/s^2

Since the blocks are moving together, the force that the 6.00-kg block exerts on the 4.00-kg block must be equal in magnitude and opposite in direction to the force that the 4.00-kg block exerts on the 6.00-kg block. Using Newton's third law, we know that these forces are equal in magnitude. Therefore:

F = 6.00 kg (2.00 m/s^2) = 12.0 N

So the magnitude of the force that the 6.00-kg block exerts on the 4.00-kg block is 12.0 N, which is closest to choice C, 8.00 N.

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