## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(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.