91影视

We have,

Mass $m=75$kg

The skis thrust of $F_{thrust}=200$N

Kinetic friction on snow is ${渭}_k=0.10$

time$t=10s$

First, we find the acceleration

$$\begin{gathered} \underset{}{鈭�}{F}_x=F_{net} \\ {F}_{thrust}-f_f=F_{net} \\ {F}_{thrust}-{渭}_{k}mg=ma \\ a=\frac{F_{thrust}-u_{k}mg}{75kg} \\ a=\frac{200-0.10\left(75\right)(9.81)}{75} \\ a=1.69m/s^2 \end{gathered}$$

Now, find the top speed

$$\begin{gathered} v_f=v_i+at \\ {v}_f=0+(1.69)脳\left(10s\right) \\ {v}_f=16.90m/s \end{gathered}$$since$v_i=0$

The ski exert a 200N of thrust force and cause an acceleration of $a=1.69$m/s² for $t=10$s and reaching a top speed of $v_f=16.90$m/s

The force acting on the ski after it accelerate is the kinetic energy, resolving forces under equilibrium, we got

$$\begin{gathered} \underset{}{鈭�}{F}_y=0 \\ N-mg=0 \end{gathered}$$

$N=mg$ (1)

Substitute Equation(1) in (2)

$f_k={渭}_{k}mg$ (2)

$$\begin{gathered} \underset{}{鈭�}{F}_x=0 \\ {F}_{net}-f_k=0 \end{gathered}$$

Let compute for the acceleration, substitute Equation (2) to above equation

localid="1649522800356" $$\begin{gathered} ma-{渭}_{k}mg=0 \\ a=\frac{{渭}_{k}mg}{m} \\ a={渭}_{k}g \\ a=0.10脳9.81 \\ a=0.98m/s^2 \end{gathered}$$

$$\begin{gathered} {v_f}^2={v_i}^2+2a鈭�x \\ {v_f}^2={v_i}^2+2a{x}_2 \\ {x}_2=\frac{{v_f}^2-{v_i}^2}{2a} \\ {x}_2=\frac{(16.90)-\left(0\right)}{2(0.90)} \\ {x}_2=145.7m \end{gathered}$$

$鈭�x$is only the distance traveled when the ski is coasting

localid="1649522949011" $$\begin{gathered} x_f=x_0+v_{0}t+\frac{1}{2}a{t}^2 \\ {x}_1=0+0\left(10\right)+\frac{1}{2}(1.69){\left(10\right)}^2=84.5m \end{gathered}$$

Total distance traveled

$$\begin{gathered} x_{total}=x_1+x_2 \\ {x}_{total}=84.5+145.7 \\ {x}_{total}=230.2m \end{gathered}$$