#### Answer

(a) The top speed is 16.9 m/s
(b) Sam travels a total distance of 230 meters.

#### Work Step by Step

(a) We can find the acceleration during the first 10 seconds.
$\sum F = ma$
$F_{thrust} - F_f = ma$
$F_{thrust} - mg~\mu_k = ma$
$a = \frac{F_{thrust} - mg~\mu_k}{m}$
$a = \frac{200~N - (75~kg)(9.80~m/s^2)(0.10)}{75~kg}$
$a = 1.687~m/s^2$
We can find the speed at the end of the acceleration period which ends at $t = 10~s$.
$v = a~t$
$v = (1.687~m/s^2)(10~s)$
$v = 16.9~m/s$
The top speed is 16.9 m/s
(b) We can find the distance $x_1$ during the 10-second acceleration period.
$x_1 = \frac{1}{2}at^2$
$x_1 = \frac{1}{2}(1.687~m/s^2)(10~s)^2$
$x_1 = 84.35~m$
We can find the rate of deceleration caused by friction after the thrust is removed.
$F_f = ma$
$mg~\mu_k = ma$
$a = g~\mu_k$
$a = (9.80~m/s^2)(0.10)$
$a = 0.98~m/s^2$
We can find the time it takes for friction to bring Sam to a stop.
$t = \frac{v_f-v_0}{a}$
$t = \frac{0-16.9~m/s}{-0.98~m/s^2}$
$t = 17.2~s$
We can find the distance $x_2$ while friction brings Sam to a stop.
$x_2 = v_0~t+\frac{1}{2}at^2$
$x_2 = (16.9~m/s)(17.2~s)-\frac{1}{2}(0.98~m/s^2)(17.2~m/s)^2$
$x_2 = 145.7~m$
We can find the total distance $x$ that Sam travels.
$x = x_1+x_2$
$x = 84.35~m+145.7~m$
$x = 230~m$
Sam travels a total distance of 230 meters.