Chapter 6, Problem 16P

To determine

The work done by gravity, work done by the normal force, the work done by the friction, and the magnitude of frictional force.

Answer to Problem 16P

The work done by the normal force is zero, the work done by gravity is $\underset{\_}{6087.42\text{J}}$, work done by the frictional force is $\underset{\_}{-2337.42\text{J}}$, and the magnitude of frictional force is $\underset{\_}{73.04\text{N}}$.

Explanation of Solution

The work done by the normal force on the skier is zero since the normal force is perpendicular to the direction of motion of the skier.

Write the expression for work done by the gravity.

    $W_{\text{grav}}=-mg\Delta y$        (I)

Here, $W_{\text{grav}}$ is the work done by the gravity, $m$ is the mass, $g$ is the acceleration due to gravity, $\Delta y$ is the vertical distance.

From Figure 1 the vertical distance is given by

    $\begin{array}{c}\Delta y=0-\left(32.0\text{m}\right)\sin \theta \\ =-\left(32.0\text{m}\right)\sin \theta \end{array}$

Use the above equation in equation (I).

    $W_{\text{grav}}=mg\left(32.0\text{m}\right)\sin \theta$        (II)

The total work done on the skier is equal to sum of work done by the gravity, and work done by the friction.

    $W_{\text{Total}}=W_{\text{grav}}+W_{\text{friction }}$

Here, $W_{\text{Total}}$ is the total work done on the skier, $W_{\text{grav}}$ is the work done by the gravity, and $W_{\text{friction }}$ is the work done by the friction.

The total work done by the skier is equal to the change in kinetic energy of the skier. The above equation is written as

    $\frac{1}{2}m{v}^2=W_{\text{grav}}+W_{\text{friction }}$

Use equation (I) in the above equation and rearrange.

    $\begin{array}{c}\frac{1}{2}m{v}^2=-mg\Delta y+W_{\text{friction }}\\ W_{\text{friction }}=\frac{1}{2}m{v}^2+mg\Delta y\end{array}$

Since $\Delta y$ is equal to $-\left(32.0\text{m}\right)\sin \theta$ the above equation is reduced to

    $W_{\text{friction }}=\frac{1}{2}m{v}^2-mg\left(32.0\text{m}\right)\sin \theta$        (III)

Write the expression for work done by frictional force.

    $W_{\text{fric}}=f_{\text{k}}\Delta x$        (IV)

Here, $W_{\text{fric}}$ is the work done by frictional force, $f_{\text{k}}$ is the frictional force, and $\Delta x$ is the distance travelled along the incline.

Rearrange the above equation to find $f_{\text{k}}$.

    $f_{\text{k}}=\frac{W_{\text{fric}}}{\Delta x}$        (V)

Conclusion:

Substitute $75.0\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$, and $15°$ for $\theta$ in equation (II), to find $W_{\text{grav}}$.

    $\begin{array}{c}{W}_{\text{grav}}=\left(75.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(32.0\text{m}\right)\sin 15°\\ =6087.42\text{J}\end{array}$

Substitute $75.0\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$, $15°$ for $\theta$ and $10\text{m}/{\text{s}}^2$ for $v$ in equation (III), to find $W_{\text{friction}}$.

    $\begin{array}{c}{W}_{\text{friction }}=\frac{1}{2}\left(75.0\text{kg}\right){\left(10\text{m}/{\text{s}}^2\right)}^2-\left(75.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(32.0\text{m}\right)\sin 15°\\ =-2337.42\text{J}\end{array}$

Substitute $-2337.42\text{J}$ for $W_{\text{friction }}$, and $32.0\text{m}$ for $\Delta x$ in equation (V), to find $f_{\text{k}}$.

    $\begin{array}{c}{f}_{\text{k}}=\frac{-2337.42\text{J}}{32.0\text{m}}\\ =73.04\text{N}\end{array}$

Therefore, the work done by the normal force is zero, the work done by gravity is $\underset{\_}{6087.42\text{J}}$, work done by the frictional force is $\underset{\_}{-2337.42\text{J}}$, and the magnitude of frictional force is $\underset{\_}{73.04\text{N}}$.