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A \(1900-\mathrm{kg}\) car experiences a combined force of air resistance and friction that has the same magnitude whether the car goes up or down a hill at \(27 \mathrm{~m} / \mathrm{s}\). Going up a hill, the car's engine needs to produce 47 hp more power to sustain the constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal?

Short Answer

Expert verified
The hill is inclined at approximately \(4.01\degree\).

Step by step solution

01

Identify the given parameters

We are given a car with a mass of \(1900 \text{ kg}\), experiencing a constant speed of \(27 \text{ m/s}\). The additional power needed for the car to go up the hill compared to going down is \(47 \text{ hp}\). We need to find the inclination angle of the hill.
02

Convert power to SI units

Convert \(47 \text{ hp}\) of additional power to watts. We use the conversion: \(1 \text{ hp} = 746 \text{ W}\). Thus, the additional power in watts is \[47 \times 746 = 35062 \text{ W}.\]
03

Express power in terms of forces

The additional power required when going up the hill can be expressed in terms of the forces acting on the car. The power due to gravity component along the hill is given by \(P = F \cdot v\), where \(F\) is the force and \(v = 27 \text{ m/s}\).
04

Relate force to the angle of incline

The force the engine needs to overcome additionally when going up the slope is the component of gravitational force, which is \(mg \sin \theta\). So, \[P = mg \sin \theta \times v.\] To find \(\theta\), set \(P = 35062 \text{ W}\) and solve. Substitute \(m = 1900 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(v = 27 \text{ m/s}\).
05

Solve for the inclination angle

With \(P = 1900 \cdot 9.8 \cdot 27 \cdot \sin \theta = 35062\), solve for \(\sin \theta\):\[sin \theta = \frac{35062}{1900 \times 9.8 \times 27}.\]Calculate \(\sin \theta\): 1. Calculate the denominator: \(1900 \times 9.8 \times 27 = 501660\).2. Divide: \(\frac{35062}{501660} = 0.0699\).3. Take \(\arcsin (0.0699)\) to find \(\theta\).
06

Calculate the angle in degrees

Find \(\theta\) by evaluating \(\theta = \arcsin(0.0699)\) using a calculator:\(\theta \approx 4.01\degree\). This angle is the incline of the hill above the horizontal.

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Key Concepts

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

Physics Problem
Solving physics problems often involves breaking down a scenario into known quantities and equations. In this case, we are dealing with a car moving up and down an inclined plane. The main task is to find the angle of inclination of the hill. It's important to clearly identify the given parameters, such as the car's mass and velocity, as well as any additional power required to maintain motion. In the problem, the car's mass is given as 1900 kg, and it moves at a constant speed of 27 m/s. The key detail is that the car needs an extra 47 horsepower to drive up the hill compared to coming down. Understanding this requirement helps us frame the problem in terms of physics concepts like force, work, and power. When solving such problems, always start by listing known values and what you need to find, as this creates a clearer path to the solution.
Gravitational Force
Gravitational force plays a significant role in this physics problem. It is the force that the engine must overcome to move the car upwards along an inclined plane. The gravitational force acting on the car can be calculated using the equation: \[ F_g = mg, \]where
  • \( m \) is the mass of the object (in kilograms),
  • \( g \) is the acceleration due to gravity, generally approximated as 9.8 m/s².
However, because the car is moving up an incline, only a component of this gravitational force acts along the plane. This component can be calculated as: \[ F = mg \sin \theta, \]where \( \theta \) is the angle of the incline. In this case, solving for \( \theta \) involves equating the component of gravitational force with the additional power required to move up the hill, as converted from horsepower to watts. Understanding this interaction between forces is vital in solving inclined plane problems.
Power Conversion
Power conversion is an essential step in handling physics problems involving moving objects on inclined planes. The problem states that the car’s engine requires an additional 47 horsepower when climbing. To use this information in your calculations, you need to convert horsepower into watts, the standard unit of power in physics:\[ 1 \text{ hp} = 746 \text{ W} \Rightarrow 47 \text{ hp} = 47 \times 746 = 35062 \text{ W} \]This conversion is crucial since power sets the stage for our calculations and reveals how much work per unit time the engine does to overcome gravitational force. Always remember to convert all units appropriately to maintain consistency in your equations. By understanding and converting power correctly, you align with physics standards, making it easier to solve problems efficiently.
Trigonometry in Physics
Trigonometry allows us to connect angles to side lengths in triangles, which is particularly useful in problems involving inclined planes. In this exercise, trigonometry helps us relate the angle of incline (\( \theta \)) to components of force. When a car moves uphill, only a portion of the gravitational force acts along the incline, directly calculated as \( mg \sin \theta \). This component equilibrates with the additional power needed (in watts). Once we set up this equation,\[ P = mg \sin \theta \times v, \]we can isolate \( \theta \) by solving for \( \sin \theta \) and use the inverse sine function to find the angle in degrees. As shown, \[ \sin \theta = \frac{P}{m \cdot g \cdot v}, \]where we've already calculated all terms, yielding \( \sin \theta \approx 0.0699 \). Using a calculator, \( \theta = \arcsin(0.0699) \approx 4.01^\circ \). Trigonometry thus bridges numerical values with physical dimensions, providing precise insights into naturally occurring phenomena.

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Most popular questions from this chapter

A \(2.00\) -kg rock is released from rest at a height of \(20.0 \mathrm{~m}\). Ignore air resistance and determine the kinetic energy, gravitational potential energy, and total mechanical energy at each of the following heights: \(20.0,10.0\), and \(0 \mathrm{~m}\).

Multiple-Concept Example 5 reviews many of the concepts that play a role in this problem. An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of \(25.0^{\circ}\) with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200 . She coasts down a distance of \(10.4 \mathrm{~m}\) before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is \(3.50 \mathrm{~m}\) below the edge. How fast is she going just before she lands?

A truck is traveling at \(11.1 \mathrm{~m} / \mathrm{s}\) down a hill when the brakes on all four wheels lock. The hill makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is \(0.750\). How far does the truck skid before coming to a stop?

One kilowatt hour (kWh) is the amount of work or energy generated when one kilowatt of power is supplied for a time of one hour. A kilowatt hour is the unit of energy used by power companies when figuring your electric bill. Determine the number of joules of energy in one kilowatt hour.

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. (a) What type of energy is changing? (b) Is the work being done by the net external force acting on the skier positive, zero, or negative? Why? (c) How is this work related to the change in the energy of the skier? A 70.3 -kg water-skier has an initial speed of \(6.10 \mathrm{~m} / \mathrm{s}\). Later, the speed increases to \(11.3 \mathrm{~m} / \mathrm{s}\). Determine the work done by the net external force acting on the skier.

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