Answer
(a) $\mu_s = 0.710$
$\mu_k = 0.472$
(b) $F = 258~N$
(c) (i) A force of 51.8 N would cause the box to move.
(ii) $a = 4.97~m/s^2$
Work Step by Step
(a) static friction:
$F_f = 313~N$
$mg~\mu_s = 313~N$
$\mu_s = \frac{313~N}{(45.0~kg)(9.80~m/s^2)}$
$\mu_s = 0.710$
kinetic friction:
$F_f = 208~N$
$mg~\mu_k = 208~N$
$\mu_k = \frac{208~N}{(45.0~kg)(9.80~m/s^2)}$
$\mu_k = 0.472$
(b) $\sum F = ma$
$F - F_f = ma$
$F = ma+F_f = (45.0~kg)(1.10~m/s^2)+208~N$
$F = 258~N$
(c) (i) static friction:
$F_f = (45.0~kg)(1.62~m/s^2)(0.710)$
$F_f = 51.8~N$
A force of 51.8 N would cause the box to move.
(ii) $\sum F = ma$
$F - F_f = ma$
$a = \frac{F-F_f}{m} = \frac{258~N-(45.0~kg)(1.62~m/s^2)(0.472)}{45.0~kg}$
$a = 4.97~m/s^2$
