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Chapter 8: Problem 72

(II) \(\mathrm{A} 75\) -kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a \(23^{\circ}\) hill. The skier is pulled a distance \(x=220 \mathrm{m}\) 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 \(\mu_{k}=0.10,\) what horsepower engine is required if 30 such skiers \((\max )\) are on the rope at one time?

Short Answer

Expert verified
The engine requires approximately 50 horsepower to pull 30 skiers.

Step by step solution

01

Identify the Forces Acting on the Skier

The forces acting on the skier include gravitational force, normal force, frictional force, and the force exerted by the rope. We need to calculate the component of gravitational force along the slope and the frictional force.
02

Calculate the Component of Gravitational Force

The component of the gravitational force acting along the slope is given by \( F_{gravity} = mg \sin(\theta) \), where \( m = 75 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( \theta = 23^{\circ} \).Substitute the values to get: \( F_{gravity} = 75 \times 9.8 \times \sin(23^\circ) \).
03

Calculate the Frictional Force

The frictional force can be calculated using \( F_{friction} = \mu_k N \), where \( \mu_k = 0.10 \) and \( N = mg \cos(\theta) \).This gives us: \( F_{friction} = 0.10 \times (75 \times 9.8 \times \cos(23^\circ)) \).
04

Calculate the Total Force Exerted by the Rope

The total force exerted by the rope is the sum of the forces needed to overcome gravity and friction:\( F_{rope} = F_{gravity} + F_{friction} \).
05

Calculate the Work Done by the Rope

The work done by the rope in pulling the skier up the hill is calculated as:\( W = F_{rope} \times x \), where \( x = 220 \text{ m} \) is the distance along the incline.
06

Convert Time to Seconds

The time taken to pull the skier to the top is 2.0 minutes. Convert this to seconds:2.0 minutes \( = 2.0 \times 60 = 120 \text{ seconds} \).
07

Calculate Power Required for One Skier

Power is calculated as:\( P = \frac{W}{t} \), where \( W \) is the work done and \( t = 120 \text{ s} \) is the time in seconds.
08

Convert to Horsepower

Convert the power from watts to horsepower:1 horsepower \( = 746 \text{ watts} \).The power for one skier in horsepower is:\( P_{hp} = \frac{P}{746} \).
09

Calculate Total Horsepower for 30 Skiers

Since there are 30 skiers, multiply the horsepower found for one skier by 30:\( P_{total} = 30 \times P_{hp} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Friction
When the skier moves up the hill, they experience friction between their skis and the snow. This friction is
  • a resistive force that tries to hinder the skier's movement.
  • For our problem, the type of friction involved is kinetic friction because the skis are sliding over the snow.

The kinetic friction force (\( F_{friction} \) in this scenario) can be determined when knowing:
  • the normal force (\( N \)) which is affected by the skier's weight and incline of the hill, and
  • the coefficient of kinetic friction (\( \mu_k \)). This indicates how easily the ski slides over snow.

Using the formula:\( F_{friction} = \mu_k \times N \), the frictional force can be evaluated. Here, as the hill has a specific incline angle (\( \theta \)), the normal force is altered to account for this inclination: \( N = mg \cos(\theta) \), where \( m \) is the mass of the skier and \( g \) is the acceleration due to gravity.
Understanding these aspects of friction helps in calculating precisely the extra effort needed to ascend the slope.
Gravitational Force
One of the most significant forces at play, especially in this incline scenario, is the gravitational force. This force directly impacts the skier moving uphill.
When an object is on an incline, gravity not only pulls it down but also exerts a component force along the slope. For the skier, this component of gravitational force is calculated using:
  • the formula \( F_{gravity} = mg\sin(\theta) \).

Parameters include:
  • \( m \): mass of the skier, measured in kilograms, influencing how strongly gravity acts on the skier.
  • \( g \): gravitational acceleration, typically \( 9.8 \) m/s\(^2\), a constant value indicating the strength of the gravitational pull.
  • \( \theta \): the angle of the incline, which impacts how much of the gravitational force acts along the slope.

Without overcoming this gravitational component, the skier would slide downhill rather than traveling upwards. This is why calculating and counteracting it is crucial for the upward movement supported by the rope.
Power
In this exercise, power is essential for understanding how the skier is pulled up at a consistent speed. Power is the rate at which work is done or energy is transferred.
  • The work (\( W \)) is done by the rope to pull the skier along the incline.

Power is calculated using the formula:
  • \( P = \frac{W}{t} \)
where:
  • \( W \) is the work done, a product of force and distance (with force being the sum of gravitational and frictional forces in this context).
  • \( t \) is the time over which this work is performed.

After figuring out how much work is required for one skier, we can convert power into horsepower, knowing that 1 horsepower (\( hp \)) equals 746 watts. Multiplying this value by the number of skiers gives the total horsepower needed.
Understanding power in this way illustrates how significantly more energy is needed to move multiple skiers up the hill as compared to just one skier, reflecting on the engine's capability to sustain this demand.

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Most popular questions from this chapter

Proper design of automobile braking systems must account for heat buildup under heavy braking. Calculate the thermal energy dissipated from brakes in a \(1500-\mathrm{kg}\) car that descends a \(17^{\circ}\) hill. The car begins braking when its speed is \(95 \mathrm{~km} / \mathrm{h}\) and slows to a speed of \(35 \mathrm{~km} / \mathrm{h}\) in a distance of \(0.30 \mathrm{~km}\) measured along the road.

(II) A ski starts from rest and slides down a \(28^{\circ}\) incline \(85 \mathrm{~m}\) long. \((a)\) If the coefficient of friction is \(0.090,\) what is the ski's speed at the base of the incline? \((b)\) If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level? Use energy methods.

(II) Show that the escape velocity for any satellite in a circular orbit is \(\sqrt{2}\) times its velocity.

Electric energy units are often expressed in the form of "kilowatt-hours." (a) Show that one kilowatt-hour (kWh) is equal to \(3.6 \times 10^{6} \mathrm{~J} .\) (b) If a typical family of four uses electric energy at an average rate of \(580 \mathrm{~W}\), how many \(\mathrm{kWh}\) would their electric bill show for one month, and (c) how many joules would this be? ( \(d\) ) At a cost of \(\$ 0.12\) per \(\mathrm{kWh}\), what would their monthly bill be in dollars? Does the monthly bill depend on the rate at which they use the electric energy?

(II) An outboard motor for a boat is rated at 55 hp. If it can move a particular boat at a steady speed of \(35 \mathrm{~km} / \mathrm{h}\), what is the total force resisting the motion of the boat?
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