Chapter 4, Problem 170P

(a)

To determine

The speed of the skier at the bottom of the slope.

Answer to Problem 170P

The speed of the skier at the bottom of the slope is $23\text{m/s}$.

Explanation of Solution

The mass of the skier is $63\text{kg}$, the length of the icy slope is $50\text{m}$, the angle between the slope and the horizontal is $32°$, the distance covered by the skier before come to rest in horizontal path is $140\text{m}$.

Diagram for skier in the slope.

Write the expression to find the total force in x-direction.

$\begin{array}{c}{\displaystyle \sum F_x}=m{a}_x\\ mg\sin \theta =m{a}_x\end{array}$

Here, $m$ is the mass of the skier, $a_x$ is the acceleration of the skier, $g$ is acceleration due to gravity, $\theta$ is the angle of the slope with horizontal.

Re-arrange the expression to find the acceleration of the skier.

$a_x=g\sin \theta$

Write the expression to find the speed of the skier at the bottom of the slope.

$v_{\text{fx}}^2-v_{\text{ix}}^2=2a_x\Delta x$

Here, $v_{\text{fx}}$ is the final velocity in x-direction, $v_{\text{ix}}$ is the initial velocity in x-direction, $\Delta x$ is the length of the slope.

The initial velocity of the skier is zero.

Substitute $0\text{m/s}$ for $v_{\text{ix}}$ to find the velocity of the skier at the bottom of the slope.

$\begin{array}{c}{v}_{\text{fx}}^2-0=2g\sin \theta \Delta x\\ v_{\text{fx}}=\sqrt{2g\sin \theta \Delta x}\end{array}$ (1)

Conclusion:

Substitute $9.8{\text{m/s}}^{\text{2}}$ for $g$, $32°$ for $\theta$ and $50\text{m}$ for $\Delta x$ to find the velocity of the skier at the bottom of the slope.

$\begin{array}{c}{v}_{\text{fx}}=\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\sin 32°\left(50\text{m}\right)}\\ =23\text{m/s}\end{array}$

(b)

To determine

The coefficient of kinetic friction between the skier and the horizontal surface.

Answer to Problem 170P

The coefficient of kinetic friction between the skier and the horizontal surface is $0.19$.

Explanation of Solution

The mass of the skier is $63\text{kg}$, the length of the icy slope is $50\text{m}$, the angle between the slope and the horizontal is $32°$, the distance covered by the skier before come to rest in horizontal path is $140\text{m}$.

Diagram for skier in horizontal surface.

Write the expression to find the acceleration on the horizontal path.

$\begin{array}{c}{v}_{\text{fx}}^2-v_{\text{ix}}^2=2a_x\Delta x\\ a_x=\frac{v_{\text{fx}}^2-v_{\text{ix}}^2}{2\Delta x}\end{array}$

Here, $v_{\text{fx}}$ is the final velocity in x-direction, $v_{\text{ix}}$ is the initial velocity in x-direction, $\Delta x$ is the length of the slope.

The final velocity of the skier is zero since he is coming to rest.

Substitute equation (1) in above expression to find the acceleration on the horizontal path.

$\begin{array}{c}{a}_x=\frac{0\text{m/s}-{\left(\sqrt{2g\sin \theta \Delta x_s}\right)}^2}{2\Delta x}\\ =-\frac{2g\sin \theta \Delta x_s}{2\Delta x}\\ =-\frac{g\sin \theta \Delta x_s}{\Delta x}\end{array}$ (2)

Here, $\Delta x_s$ is the length of the slope.

Write the expression to find the total force in x-direction.

$\begin{array}{c}{\displaystyle \sum F_x}=m{a}_x\\ -f_k=m{a}_x\\ -\left({\mu }_{k}N\right)=m{a}_x\\ {\mu }_k=-\frac{m{a}_x}{N}\end{array}$ (3)

Here, $m$ is the mass of the skier, $a_x$ is the acceleration of the skier, $g$ is acceleration due to gravity.

Write the expression to find the total force in y-direction.

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ N-mg=0\\ N=mg\end{array}$ (4)

Substitute equations (2), (4) in (3) to find the coefficient of kinetic friction between the skier and the horizontal surface.

$\begin{array}{c}{\mu }_k=-\frac{m\left(-\frac{g\sin \theta \Delta x_s}{\Delta x}\right)}{mg}\\ =\frac{\sin \theta \Delta x_s}{\Delta x}\end{array}$

Conclusion:

Substitute $32°$ for $\theta$, $140\text{m}$ for $\Delta x$ and $50\text{m}$ for $\Delta x_s$ to find the coefficient of kinetic friction between the skier and the horizontal surface.

$\begin{array}{c}{\mu }_k=\frac{\sin 32°\left(50\text{m}\right)}{140\text{m}}\\ =0.19\end{array}$