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 two blocks are in contact on a frictionless surface, they will have the same acceleration. Using Newton's Second Law, we can find the acceleration of the system: 
$\Sigma F_{total} = (6.00 kg + 4.00 kg) \cdot a$
$20.0 N - F_{6on4} = 10.0 kg \cdot a$
$F_{6on4} = 20.0 N - 10.0 kg \cdot a$
We still need to find the acceleration $a$, which we can get from either of the blocks using Newton's Second Law again: 
$\Sigma F_{6} = 6.00 kg \cdot a$
$20.0 N = 6.00 kg \cdot a$
$a = 3.33 m/s^2$
Substituting this into our equation for $F_{6on4}$, we get: 
$F_{6on4} = 20.0 N - 10.0 kg \cdot 3.33 m/s^2 = 8.01 N \approx 8.00 N$ 
Therefore, the magnitude of the force that the 6.00-kg block exerts on the 4.00-kg block is 8.00 N, or choice C.

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