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Chapter 13: Problem 2

While a \(1700-\mathrm{kg}\) automobile is moving at a constant speed of \(15 \mathrm{~m} / \mathrm{s}\), the motor supplies \(16 \mathrm{~kW}\) of power to overcome friction, wind resistance, and so on. (a) What power must the motor supply if the car is to move up an \(8.0 \%\) grade \((8.0 \mathrm{~m}\) vertically for each \(100 \mathrm{~m}\) horizontally) at \(15 \mathrm{~m} / \mathrm{s} ?\) (b) At what downgrade, expressed in percentage terms, would the car coast at \(15 \mathrm{~m} / \mathrm{s}\) ?

Short Answer

Expert verified
The power that the motor would need to supply to maintain a speed of 15 m/s while moving up an 8.0% incline is approximately 35.28 kW and the car would coast at 15 m/s at an approximately 9.4% downgrade.

Step by step solution

01

Calculate work done against gravity

To keep the power constant, we'll start by calculating the work done against gravity while moving on an incline. The work done against gravity is given by \( Work = m \cdot g \cdot h \), where \( m = 1700 \, kg \) is the mass of the car, \( g = 9.8 \, m/s^2 \) is the acceleration due to gravity, and \( h \) is the increase in height which is calculated by \((8/100) \cdot distance \) .
02

Find power required to move car uphill at constant speed

The power \( P \) required to move the car uphill at a constant speed is \( Power = Work/time = m \cdot g \cdot h / t \). Let's denote the speed of the car as \( v = 15 \, m/s \), then the time \( t \) needed to travel a certain distance \( d \) would be \( d/v \). Therefore, the total power \( P_{total} \) required for the car to move uphill at that constant speed would be the sum of power to overcome friction and wind resistance and the power against gravity \( P_{total} = P_{friction} + P_{gravity} \), where \( P_{friction} = 16 \, kW \).
03

Calculate downgrade at which the car could coast at constant speed

To find the downgrade at which the car would move at a constant speed of \( 15 \, m/s \) without any power (coasting), we would set the power required to be equal to the power to overcome friction only (since there is no extra power supplied to the car). Therefore, the height decrease would be calculated as \( h = Power_{friction} \cdot t /(m \cdot g) = Power_{friction} \cdot v /(m \cdot g) \). And this height decrease \( h \) would be equal to the downgrade \( grade = h/distance \cdot 100% \).

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Most popular questions from this chapter

A ball loses \(15.0 \%\) of its kinetic energy when it bounces back from a concrete walk. With what speed must you throw it vertically down from a height of \(12.4 \mathrm{~m}\) to have it bounce back to that same height? Ignore air resistance.

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