# Chapter 7 - Newton's Third Law - Exercises and Problems - Page 177: 12

The mass of the cable is 5.0 kg

#### Work Step by Step

Consider the system of the block and the cable. We can find the acceleration of the system. $a = \frac{v^2-v_0^2}{2x}$ $a = \frac{(4.0~m/s)^2-0}{(2)(2.0~m)}$ $a = 4.0~m/s^2$ Let $m_b$ be the mass of the block. Let $m_c$ be the mass of the cable. We can find the mass of the cable. $F = (m_b+m_c)~a$ $m_c = \frac{F - m_b~a}{a}$ $m_c = \frac{(100~N) - (20~kg)(4.0~m/s^2)}{4.0~m/s^2}$ $m_c = 5.0~kg$ The mass of the cable is 5.0 kg

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