(II) A 65-kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 23° hill. The skier is pulled a distance\(x = 320{\rm{ m}}\) along the incline and it takes 2.0 min to reach the top of the hill. If the coefficient of kinetic friction between the snow and skis is \({\mu _{\rm{k}}} = 0.10\), what horsepower engine is required if 30 such skiers (max) are on the rope at one time?

In this problem, the force along the hill exerted on the skier is parallel to the displacement of the skier.So, work done will be equal to the product of this force and distance covered.

Given data:

The mass of the skier is .\(m = 65\;{\rm{kg}}\)..

The angle made by the skier is\(\theta = 23^\circ \).

The distance covered is\(d = x = 320\;{\rm{m}}\).

The time taken to reach the top of the hill is\(t = 2\;{\rm{min}}\).

The coefficient of kinetic friction is\({\mu _{\rm{k}}} = 0.10\).

The number of skiers is\(N = 30\).

The free body diagram of the skier is as follows:

Let Fbe the force exerted on the skier.

Consider the direction along the plane as x-direction, and the perpendicular to the plane as the y-direction.

The relation of the forces in the y-direction is given by:

\(\begin{aligned}\Sigma {F_{\rm{y}}} &= 0\\{F_{\rm{n}}} - W\cos \theta &= 0\\{F_{\rm{n}}} &= mg\cos \theta \end{aligned}\)

Here, \({F_{\rm{n}}}\)is the normal force, W is the weight, and g is the gravitational acceleration.

Since the speed is constant, there is no acceleration along the plane.

The relation of forces in the x-direction is given by:

\(\begin{aligned}\Sigma {F_{\rm{x}}} &= 0\\F - W\sin \theta - {F_{\rm{f}}} &= 0\\F &= mg\sin \theta + {\mu _{\rm{k}}}{F_{\rm{n}}}\end{aligned}\)

Here,\({F_{\rm{f}}}\)is the frictional force.

On plugging the values in the above relation, you get:

\(F = mg\sin \theta + {\mu _{\rm{k}}}mg\cos \theta \)

The relation for finding the horsepower of the engine for N number of skiers is given by:

\(\begin{aligned}P &= N\frac{W}{t}\\P &= N\left( {\frac{{F \times d}}{t}} \right)\end{aligned}\)

On plugging the values in the above relation, you get:

\(\begin{aligned}P &= N\left( {\frac{{mg\sin \theta + {\mu _{\rm{k}}}mg\cos \theta }}{t}} \right)d\\P &= \left( {30} \right)\left( {\frac{{\left( {65\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\sin 23^\circ + \left( {0.10} \right)\left( {65\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\cos 23^\circ }}{{2\;{\rm{min}} \times \frac{{60\;{\rm{s}}}}{{1\;{\rm{min}}}}}}} \right)\left( {320\;{\rm{m}}} \right)\\P &= \left( {24627.2\;{\rm{W}} \times \frac{{1\;{\rm{hp}}}}{{746\;{\rm{W}}}}} \right)\\P &= 33.01\;{\rm{hp}}\\P &\approx 33{\rm{ hp}}\end{aligned}\)

Thus, \(P = 33{\rm{ hp}}\) is the required horsepower.