Chapter 6, Problem 16P

To determine

The work done by gravity and normal force on the skier.

Answer to Problem 16P

The work done by gravity on the skier is $6.09\text{kJ}$ and work done by normal force on the skier is $0\text{kJ}$.

Explanation of Solution

Write the expression for work done.

$W_g=Fx\cos \theta$

Here, $W_g$ is the work done by gravity, F is the force, x is the displacement and $\theta$ is the angle force and displacement.

Write the expression for force.

$F=mg$

Here, m is the mass and g is the acceleration due to gravity.

Replace F by mg in the expression for $W_g$.

$W_g=mgx\cos \theta$

The angle force and displacement is,

$\begin{array}{c}\theta =90°-15°\\ =75°\end{array}$

Conclusion:

Substitute 75 kg for m, 32.0 m for x, $9.8{\text{m/s}}^{\text{2}}$ for g and $75°$ for $\theta$ to get $W_g$.

$\begin{array}{c}{W}_g=\left(75\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(32.0\text{m}\right)\cos 75°\\ =6087\text{J}\\ \approx \left(\text{6087}\text{J}\right)\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ =6.09\text{kJ}\end{array}$

Normal force is perpendicular to the direction of motion. Hence, value of $\theta$ is $90°$. Therefore, work done by normal force is zero since cosine of $90°$ vanishes to zero.

Therefore, work done by gravity on the skier is $6.09\text{kJ}$ and work done by normal force on the skier is $0\text{kJ}$.

To determine

The work done by friction on the skier.

Answer to Problem 16P

The work done by friction on the skier is $-\text{2}\text{.34}\text{kJ}$.

Explanation of Solution

Write the expression to find the total work done.

$W_{\text{tot}}=W_g+W_f$ (I)

Here, $W_f$ is the work done by friction, $W_{\text{tot}}$ is the total work done.

The total work done accounts for the change in kinetic energy. The initial kinetic energy of the skier is zero.

$\begin{array}{c}{W}_{\text{tot}}=\Delta K\\ =\frac{1}{2}m{v}^2-0\\ =\frac{1}{2}m{v}^2\end{array}$

Substitute equation (I) in expression to find the work done by friction on the skier.

$\begin{array}{c}{W}_g+W_f=\frac{1}{2}m{v}^2\\ W_f=\frac{1}{2}m{v}^2-W_g\end{array}$

Conclusion:

Substitute $6.09\text{kJ}$ for $W_g$, 75 kg for m, $10.0\text{m/s}$ for $v$ to find the work done by friction on the skier.

$\begin{array}{c}{W}_f=\frac{1}{2}\left(75\text{kg}\right){\left(10.0\text{m/s}\right)}^2-6.09\text{kJ}\\ \text{=}-\text{2}\text{.34}\text{kJ}\end{array}$

To determine

The force of friction.

Answer to Problem 16P

The force of friction is $73.0\text{N}$.

Explanation of Solution

Write the expression to find the force of friction.

$\begin{array}{c}{W}_f=f_{k}x\cos 180°\\ f_k=-\frac{W_f}{x}\end{array}$

Conclusion:

Substitute $-\text{2}\text{.34}\text{kJ}$ for $W_f$ and $32.0\text{m}$ for $x$ to find the force of friction.

$\begin{array}{c}{f}_k=-\frac{-\text{2}\text{.34}\text{kJ}}{32.0\text{m}}\\ =73.0\text{N}\end{array}$

To determine

The coefficient of kinetic friction.

Answer to Problem 16P

The coefficient of kinetic friction is $0.103$.

Explanation of Solution

The diagram for skier.

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

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

Here, $N$ is the normal force, $\theta$ is the angle between weight and y-axis.

Write the expression to find the frictional force.

$f_k={\mu }_{k}N$

Here, $f_k$ is the frictional force, ${\mu }_k$ is the coefficient of kinetic friction.

Re-arrange the expression to find the coefficient of kinetic friction.

${\mu }_k=\frac{f_k}{N}$

Substitute equation (II) in the expression to find the coefficient of kinetic friction.

${\mu }_k=\frac{f_k}{mg\cos \theta }$

Conclusion:

Substitute $73.0\text{N}$ for $f_k$, $75.0\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$ and $15.0°$ for $\theta$ to find the coefficient of kinetic friction.

$\begin{array}{c}{\mu }_k=\frac{73.0\text{N}}{\left(75.0\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\cos 15.0°}\\ =0.103\end{array}$