Answer
(a) $F = 32.9~N$
(b) The box slides a distance of 3.13 meters before coming to rest.
Work Step by Step
(a) To keep the box moving at a constant speed, the worker needs to exert a horizontal force that is equal in magnitude to the force of kinetic friction.
$F = mg~\mu_k = (16.8~kg)(9.80~m/s^2)(0.20)$
$F = 32.9~N$
(b) We can find the rate of deceleration of the box.
$F_f = ma$
$a = \frac{F_f}{m}= \frac{mg~\mu_k}{m} = g~\mu_k$
$a = (9.80~m/s^2)(0.20) = 1.96~m/s^2$
We can find the distance the box slides before coming to rest.
$x = \frac{v^2-v_0^2}{2a} = \frac{0-(3.50~m/s)^2}{(2)(-1.96~m/s^2)}$
$x = 3.13~m$
The box slides a distance of 3.13 meters before coming to rest.
