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 only forces acting on the blocks are their weight and the applied force. Since the blocks are in contact with each other, they exert forces on each other with magnitudes that are equal and opposite by Newton's Third Law. Therefore, the force that the 6.00-kg block exerts on the 4.00-kg block has the same magnitude as the 20.0-N applied force. However, the 6.00-kg block also experiences an internal force due to the interaction with the 4.00-kg block. Using Newton's Second Law, we can write
F_applied - F_internal = ma
where F_applied is the applied force, F_internal is the internal force exerted by the 4.00-kg block on the 6.00-kg block, and a is the acceleration of the system. Since the surface is frictionless, the blocks move together with the same acceleration. Thus, we can write
a = F_applied / (m1 + m2) = 20.0 N / 10.0 kg = 2.00 m/s^2
Substituting this into the previous equation, we have
F_internal = m1 a - F_applied = (6.00 kg)(2.00 m/s^2) - 20.0 N = 8.00 N
Therefore, the magnitude of the force that the 6.00-kg block exerts on the 4.00-kg block is 8.00 N, which is choice (C).

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