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Automotive Power II. (a) If 8.00 hp are required to drive a \(1800-\) -kg automobile at 60.0 \(\mathrm{km} / \mathrm{h}\) on a level road, what is the total retarding force due to friction, air resistance, and so on? (b) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) up a 10.0\(\%\) grade (a hill rising 10.0 \(\mathrm{m}\) vertically in 100.0 \(\mathrm{m}\) horizon- tally)? (c) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) down a 1.00\(\%\) grade? (d) Down what percent grade would the car coast at 60.0 \(\mathrm{km} / \mathrm{h}\) ?

Short Answer

Expert verified
(a) 357.8 N (b) 47.4 hp (c) 4.06 hp (d) 2.03% grade.

Step by step solution

01

Understanding the Power Conversion

1 horsepower (hp) is equivalent to 746 Watts (W). Thus, 8 hp can be converted to Watts as follows: \[8 \text{ hp} = 8 \times 746 \text{ W} = 5968 \text{ W}\] This is the required power to keep the car moving at 60.0 km/h on a level road.
02

Calculating Retarding Force

The power formula is given by \( P = F \times v \), where \( F \) is the force and \( v \) is the velocity in m/s. First, convert 60.0 km/h to m/s: \[v = \frac{60.0 \text{ km/h} \times 1000 \text{ m/km}}{3600 \text{ s/h}} = 16.67 \text{ m/s} \] Solve for \( F \): \[ F = \frac{P}{v} = \frac{5968 \text{ W}}{16.67 \text{ m/s}} \approx 357.8 \text{ N} \] This is the total retarding force due to friction and air resistance.
03

Power for Uphill Driving

A 10.0% grade means the hill rises 10 m for every 100 m. The additional force required to overcome gravity is \( F_{g} = m \cdot g \cdot \sin(\theta) \). With \( \sin(\theta) \approx 0.1 \) for small angles: \[ F_{g} = 1800 \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.1 = 1764 \text{ N}\] Total force \(F_{\text{total}} = 357.8 \text{ N} + 1764 \text{ N} = 2121.8 \text{ N} \). Required power: \[ P = F_{\text{total}} \times v = 2121.8 \text{ N} \times 16.67 \text{ m/s} = 35374 \text{ W} \] Convert to hp: \[ \frac{35374 \text{ W}}{746 \text{ W/hp}} \approx 47.4 \text{ hp}\]
04

Power for Downhill Driving

For a 1.00% grade, the additional force due to gravity is \( F_{g} = 1800 \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.01 \approx 176.4 \text{ N} \). Net retarding force \( F_{\text{net}} = 357.8 \text{ N} - 176.4 \text{ N} = 181.4 \text{ N} \). Required power: \[ P = 181.4 \text{ N} \times 16.67 \text{ m/s} = 3024 \text{ W}\] Convert to hp: \[ \frac{3024 \text{ W}}{746 \text{ W/hp}} \approx 4.06 \text{ hp} \]
05

Calculating Grade for Coasting

When coasting, the power is zero, meaning the gravitational force equals retarding force. Using \( F_{g} = 357.8 \text{ N} \), solve for \( \sin(\theta) \): \[ \sin(\theta) = \frac{F_{g}}{m \cdot g} = \frac{357.8 \text{ N}}{1800 \text{ kg} \cdot 9.8 \text{ m/s}^2} \approx 0.0203 \] Converting to percent: \( 0.0203 \times 100 \approx 2.03 \text{%} \).

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Key Concepts

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

Power Conversion
Power conversion is the fundamental step in understanding how much work a machine, such as a car, can do based on its power source. It involves converting units of horsepower (hp) into watts (W), which are more commonly used in scientific calculations. Knowing that 1 horsepower is equivalent to 746 watts is essential. For example, if a car engine is using 8 hp, this translates to \(8 \times 746\) W, or 5968 W. This conversion is crucial because it provides us with a common unit that can then be used in further calculations such as force and energy needs, especially when performing tasks like going uphill or fighting drag and friction on a flat surface.
Retarding Forces
Retarding forces are the forces that work against the motion of the car, slowing it down. These include friction from the road and resistance from the air. To calculate the total retarding force when a car is cruising at a constant speed, we can use the formula \( P = F \times v \). Here, \( P \) is the power in watts, \( F \) is the force in newtons, and \( v \) is the speed in meters per second (converted from kilometers per hour). For our exercise, with a known power of 5968 W and a speed of 16.67 m/s, the retarding force comes out to about 357.8 N. This force needs to be constantly overcome by the engine to maintain a steady speed on a level road.
Uphill and Downhill Driving
Driving uphill and downhill introduces additional forces into the power calculation. When driving uphill at a 10% grade, the car has to work against gravity. Calculating this involves recognizing that the hill’s rise adds an additional force component \( F_{g} = m \cdot g \cdot \sin(\theta) \). For a 10% grade, \( \sin(\theta) \approx 0.1 \), leading to an uphill gravitational force of 1764 N. Combined with the retarding force, the total force becomes 2121.8 N. The power needed is then calculated as \( P = F_{\text{total}} \times v \), which results in a significant increase in power requirements to about 47.4 hp.
On the other hand, when driving downhill, gravity aids in propulsion. This reduces the net retarding force, as calculated for a 1% grade, resulting in lower necessary power, around 4.06 hp.
Coasting Calculations
Coasting occurs when the car moves at a steady speed without additional power from the engine. This happens when the gravitational force matches the retarding forces perfectly, requiring no extra energy from the engine. To find the downhill slope where coasting occurs, set the retarding force equal to the gravitational force component. The equation \( F_{g} = 357.8 \) N is used to solve for \( \sin(\theta) \), leading to approximately 0.0203. Converting this to a percentage gives a downhill grade of about 2.03%. This grade indicates a balance between gravity and the retarding forces, allowing the car to maintain speed without engine power.

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Most popular questions from this chapter

You apply a constant force \(\vec{\boldsymbol{F}}=(-68.0 \mathrm{N}) \hat{\boldsymbol{\imath}}+(36.0 \mathrm{N}) \hat{\boldsymbol{j}}\) to a 380 -kg car as the car travels 48.0 \(\mathrm{m}\) in a direction that is \(240.0^{\circ}\) . counterclockwise from the \(+x\) -axis. How much work does the force you apply do on the car?

CALC Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount \(x,\) a force along the \(x\) -axis with \(x\) -component \(F_{x}=k x-b x^{2}+c x^{3}\) must be applied to the free end. Here \(k=100 \mathrm{N} / \mathrm{m}, b=700 \mathrm{N} / \mathrm{m}^{2},\) and \(c=12,000 \mathrm{N} / \mathrm{m}^{3} .\) Note that \(x>0\) when the spring is stretched and \(x<0\) when it is compressed. (a) How much work must be done to stretch this spring by 0.050 m from its unstretched length? (b) How much work must be done to compress this spring by 0.050 m from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of \(F_{x}\) on \(x\) . (Many real springs behave qualitatively in the same way.)

Rescue. Your friend (mass 65.0 \(\mathrm{kg} )\) is standing on the ice in the middle of a frozen pond. There is very litle friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 s and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) whileyou remain at rest. What is the average power supplied by the force you applied?

A batter hits a baseball with mass 0.145 \(\mathrm{kg}\) straight upward with an initial speed of 25.0 \(\mathrm{m} / \mathrm{s}\) . (a) How much work has gravity done on the baseball when it reaches a height of 20.0 \(\mathrm{m}\) above the bat? (b) Use the work-energy theorem to calculate the speed of the baseball at a height of 20.0 m above the bat. You can ignore air resistance. (c) Does the answer to part (b) depend on whether the baseball is moving upward or downward at a height of 20.0 \(\mathrm{m} ?\) Explain.

CALC The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the earth's surface this force is equal to the object's normal weight \(m g,\) where \(g=9.8 \mathrm{m} / \mathrm{s}^{2},\) and at large distances, the force is zero. If a \(20,000-\mathrm{kg}\) asteroid falls to earth from a very great distance away, what will be its minimum speed as it strikes the earth's surface, and how much kinetic energy will it impart to our planet? You can ignore the effects of the earth's atmosphere.

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