## University Physics with Modern Physics (14th Edition)

$m_2 = 36.1~kg$
We can find the acceleration of the system. $y = \frac{1}{2}at^2$ $a = \frac{2y}{t^2} = \frac{(2)(12.0~m)}{(3.00~s)^2}$ $a = 2.67~m/s^2$ We can set up a force equation for each block. First we can set up a force equation for $m_1$ $\sum F = m_1~a$ $T - m_1~g~sin(\alpha) - m_1~g~cos(\alpha)~\mu_k = m_1~a$ $T = m_1~g~sin(\alpha) + m_1~g~cos(\alpha)~\mu_k + m_1~a$ $T = (20.0~kg)(9.80~m/s^2)~sin(53.1^{\circ}) + (20.0~kg)(9.80~m/s^2)~cos(53.1^{\circ})(0.40) + (20.0~kg)(2.67~m/s^2)$ $T = 257.2~N$ We can set up a force equation for $m_2$ $\sum F = m_2~a$ $m_2~g - T = m_2~a$ $m_2~(g-a) = T$ $m_2 = \frac{T}{g-a}$ $m_2 = \frac{257.2~N}{(9.80~m/s^2)-(2.67~m/s^2)}$ $m_2 = 36.1~kg$