On a horizontal frictionless surface, a block with a 4.00 kg mass in contact with a block of mass 2.00 kg, is accelerated by a +12.0 N force. Determine the force that the 2.00 kg block exerts on the 4.00 kg block.

+4 N

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-12 N

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-4 N

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+6 N

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Solution

The correct option is C
-4 N

$M a = + 12 N (M = 4 + 2)$

$a = \frac{+ 12 N}{6.00 k g} = + 2 m / s^{2}$

$F_{o n - 2 k g} = m a = (2 k g) (+ 2 m / s^{2}) = + 4 N$

$F_{o n - 4 k g} = - F_{o n - 2 k g} = - 4 N$

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On a horizontal frictionless surface, a block with a 4.00 kg mass in contact with a block of mass 2.00 kg, is accelerated by a +12.0 N force. Determine the force that the 2.00 kg block exerts on the 4.00 kg block.

The correct option is C -4 N Ma=+12N(M=4+2) a=+12 N6.00 kg=+2 m/s2 Fon−2 kg=ma=(2 kg)(+2 m/s2)=+4N Fon−4 kg=−Fon−2 kg=−4N

The correct option is C
-4 N

$M a = + 12 N (M = 4 + 2)$

$a = \frac{+ 12 N}{6.00 k g} = + 2 m / s^{2}$

$F_{o n - 2 k g} = m a = (2 k g) (+ 2 m / s^{2}) = + 4 N$

$F_{o n - 4 k g} = - F_{o n - 2 k g} = - 4 N$