Q59E - University Physics with Modern Physics

Question

A ski tow operates on a $\mathbf{15}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$ slope of length$\mathbf{300}\mathbf{ }\mathbf{m}$. The rope moves at $\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{km}\mathbf{/}\mathbf{h}$ and provides power for $\mathbf{50}$ riders at one time, with an average mass per rider of $\mathbf{70}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{kg}$. Estimate the power required to operate the tow.

Step-by-Step Solution

Answer

Slope = $15.0^{°}$

Length = $300 m$

Rope movement = $12.0 km/h$

Mass = $70. kg$ 

Step 2: Determine the power required to operate the tow

                                   

The component of weight is the net force acting on the tow.

$F_{Rider}=-mg\hspace{0.33em}sin{15}^{\circ }$.

The force component of gravity is negative, and the work done by this force is also negative. So, the rope must do positive work to lift the skiers.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{rope}}=-{\mathrm{F}}_{\mathrm{Rider}} \\ =\mathrm{mg}\hspace{0.33em}\sin {15}^{\circ } \end{gathered}$$

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

There are N riders, so, a total required force for N riders is as follows:

$F\hspace{0.33em}=Nmg\hspace{0.33em}sin{15}^{\circ }$

Write the expression for the power required to operate the row.

$\mathrm{P}=\mathrm{Fv}$

Here, m is the net force on the g riders and v is the speed of the rope.

Substitute $\mathrm{Nmg}\hspace{0.33em}{\sin }^{\circ }$for F

$\mathrm{p}=\mathrm{Nmgv} \sin \circ$

Step3 :

The speed of the rope is $12\mathrm{km}/\mathrm{h}$. Convert the units of the speed o rope from $\mathrm{km}/\mathrm{h}$ to $m/s$ 

$$\begin{gathered} \mathrm{v}=\left(12\mathrm{km}/\mathrm{h}\right)\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\left(\frac{1\mathrm{h}}{3600\hspace{0.33em}\mathrm{s}}\right) \\ =3.3\hspace{0.33em}\mathrm{m}/\mathrm{s} \end{gathered}$$

Substitute 5 for N, 70kg for m, $9.8 \mathrm{m}/{\mathrm{s}}^2$for g, and 3.3 m/s for v .

$$\begin{gathered} \mathrm{p}=\mathrm{p}=\mathrm{Nmgv} \sin {15}^{\circ } \\ =\left(50\right)\left(70\mathrm{kg}\right)(9.8\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2)(3.3\hspace{0.33em}\mathrm{m}/\mathrm{s})\sin \hspace{0.33em}{15}^{\circ } \\ =(29.6\times {10}^2\mathrm{w})\left(\frac{1\hspace{0.33em}\mathrm{kw}}{{10}^2\mathrm{w}}\right) \\ =29.6\mathrm{kw} \end{gathered}$$

Therefore, the power required to operate the tow is $29.6kw$