Answer
(a) The 2.0-kg block exerts a force of 6.0 N on the 3.0-kg block.
(b) The 2.0-kg block exerts a force of 10 N on the 1.0-kg block.
Work Step by Step
(a) We can find the acceleration of the system of all three blocks.
$F = (m_1+m_2+m_3)~a$
$a = \frac{F}{(m_1+m_2+m_3)}$
$a = \frac{12~N}{1.0~kg+2.0~kg+3.0~kg}$
$a = 2.0~m/s^2$
The force that the 2.0-kg block exerts on the 3.0-kg block provides the force to accelerate the 3.0-kg block with an acceleration of $2.0~m/s^2$. We can find the force that the 2.0-kg block exerts on the 3.0-kg block.
$F = m_3~a$
$F = (3.0~kg)(2.0~m/s^2)$
$F = 6.0~N$
The 2.0-kg block exerts a force of 6.0 N on the 3.0-kg block.
(b) The force that the 1.0-kg block exerts on the 2.0-kg block provides the force to accelerate the 2.0-kg block and the 3.0-kg block with an acceleration of $2.0~m/s^2$. We can find the force that the 1.0-kg block exerts on the 2.0-kg block.
$F = (m_2+m_3)~a$
$F = (2.0~kg+3.0~kg)(2.0~m/s^2)$
$F = 10~N$
The 1.0-kg block exerts a force of 10 N on the 2.0-kg block. By Newton's third law, the 2.0-kg block exerts an equal and opposite force on the 1.0-kg block. Therefore, the 2.0-kg block exerts a force of 10 N on the 1.0-kg block.
