Answer
(a) $F_f = 0$
(b) $F_f = 6.0~N$
(c) $F = 16.0~N$
(d) $F = 8.0~N$
(e) $F_f = 8.0~N$
$a = 2.45~m/s^2$
Work Step by Step
(a) If the box is at rest and no horizontal force is applied to the box, then the friction force is zero.
(b) If a horizontal force of 6.0 N is applied to the box, the force of static friction, which opposes this force, is 6.0 N.
(c) We can find the maximum possible force of static friction.
$F_f = F_N ~\mu_s = (40.0~N)(0.40)$
$F_f = 16.0~N$
The maximum possible force of static friction is 16.0 N. If the horizontal force is greater than 16.0 N then the box will start moving.
(d) The minimum force $F$ to keep the box moving will be equal in magnitude to the force of kinetic friction.
$F = F_N~\mu_k = (40.0~N)(0.20) = 8.0~N$
(e) We can find the mass of the box.
$40.0 ~N = mg$
$m = \frac{40.0~N}{9.80~m/s^2} = 4.08~kg$
As we found in part (d), the magnitude of the force of kinetic friction is 8.0 N. We can find the acceleration of the box.
$\sum F = ma$
$F-F_f = ma$
$a = \frac{18.0~N - 8.0~N}{4.08~kg}$
$a = 2.45~m/s^2$
