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

(a) $T = 11.4~N$ (b) $m = 2.57~kg$
(a) Let's consider the system of the 5.00-kg box. We can find the force of kinetic friction $F_f$ acting on the box. $F_f = F_N~\mu_k$ $F_f = (mg-F~sin(\theta))~\mu_k$ $F_f = ((5.00~kg)(9.80~m/s^2)-(40.0~N)~sin(53.1^{\circ}))\times (0.30)$ $F_f = 5.10~N$ We can use a force equation to find the tension $T$. $\sum F = ma$ $F_x - T - F_f = ma$ $T = F_x - F_f - ma$ $T = [40.0~N~cos(53.1^{\circ})]-(5.10~N)-(5.00~kg)(1.50~m/s^2)$ $T = 11.4~N$ (b) Let's consider the system of box A. $\sum F = ma$ $T - mg~\mu_k = ma$ $ma+ mg~\mu_k = T$ $m = \frac{T}{a+g~\mu_k}$ $m = \frac{11.4~N}{(1.50~m/s^2)+(9.80~m/s^2)(0.30)}$ $m = 2.57~kg$