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

(a) If the blocks are at rest, we can consider the system of the 2.0-kg block to find the tension in the string. Let $m_1$ be the mass of the 2.0-kg block. $\sum F = m_1~a$ $m_1~g -T = 0$ $T = m_1~g$ $T = (2.0~kg)(9.80~m/s^2)$ $T = 19.6 ~N$ (b) We can find the component of the 4.0-kg block's weight which is directed down the slope. Let $m_2$ be the mass of the 4.0-kg block. $F_{slope} = m_2~g~sin(\theta)$ $F_{slope} = (4.0~kg)(9.80~m/s^2)~sin(35^{\circ})$ $F_{slope} = 22.48~N$ Since the component of the 4.0-kg block's weight directed down the slope is greater than the weight of the 2.0-kg block, the 4.0-kg block will move down the slope. (c) We can set up a force equation for the 2.0-kg block. $\sum F = m_1~a$ $T - m_1~g = m_1~a$ $a = \frac{T - m_1~g}{m_1}$ We can use this expression in the force equation for the 4.0-kg block. Note that the magnitude of acceleration is the same for both blocks. $\sum F = m_2~a$ $m_2~g~sin(\theta) - T = m_2~a$ $m_2~g~sin(\theta) - T = m_2~(\frac{T - m_1~g}{m_1})$ $m_1~m_2~g~sin(\theta) - m_1~T = m_2~T - m_1~m_2~g$ $m_1~m_2~g~[1+sin(\theta)] = T~(m_1+m_2)$ $T = \frac{m_1~m_2~g~[1+sin(\theta)]}{m_1+m_2}$ $T = \frac{(2.0~kg)(4.0~kg)(9.80~m/s^2)~[1+sin(35^{\circ})]}{2.0~kg+4.0~kg}$ $T = 20.6~N$