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

(a) $F = \frac{mg~\mu_k}{cos(\theta)+sin(\theta)~\mu_k}$ (b) $F = 290~N$
(a) Since the speed is constant, the horizontal component of the force $F$ is equal in magnitude to the force of kinetic friction. $F~cos(\theta) = F_N~\mu_k$ $F~cos(\theta) = [mg-F~sin(\theta)]~\mu_k$ $F~cos(\theta)+F~sin(\theta)~\mu_k = mg~\mu_k$ $F = \frac{mg~\mu_k}{cos(\theta)+sin(\theta)~\mu_k}$ (b) We can use the answer in part (a) to find the required force. $F = \frac{mg~\mu_k}{cos(\theta)+sin(\theta)~\mu_k}$ $F = \frac{(90~kg)(9.80~m/s^2)(0.35)}{cos(25^{\circ})+sin(25^{\circ})(0.35)}$ $F = 290~N$