Question

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

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

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 \((8/100) \cdot distance \) .

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 \( P_{friction} = 16 \, kW \).

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 \( grade = h/distance \cdot 100% \).

Key concepts

Power and Work

When moving objects, we often talk about the concepts of power and work, which are vital in understanding forces in physics. Work is done when a force moves an object through a distance. The formula for work is:

• \( Work = Force \, \times \, Distance \)

In the context of moving a car uphill, the force at work is the force of gravity acting on the car's mass as it travels upward.

Power, on the other hand, is the rate at which work is done. It is measured in watts (W). If more power is needed, it means the work is being done faster or more forcefully. In our car scenario, it requires more power to overcome gravity on an incline than on a flat road.

The relationship between power, work, and time can be expressed in the formula:

• \( Power = \frac{Work}{Time} \)

Therefore, to calculate the power needed to move the car uphill, we first calculate the work done against gravity, and then divide it by the time it takes to travel a certain distance.

Friction and Wind Resistance

Friction and wind resistance act as opposing forces against the movement of the car. These forces need to be overcome to maintain constant velocity. Friction is the resistance force from the surfaces in contact, in this case, the car's tires and the road. Wind resistance refers to the force that the air exerts on the moving vehicle, slowing it down.

In our scenario, the car's motor supplies a continuous power of 16 kW just to counteract these forces while moving on a level surface. This ensures that the car maintains a steady speed as energy is continuously used to overcome these resistances.

When moving uphill, additional power is necessary, since gravity adds another dimension of resistance. While moving downhill, the forces of friction and wind resistance aid in slowing the car, which would potentially allow it to maintain speed with less engine power or even coast without any power.

Inclined Planes

An incline or slope adds complexity to motion by introducing a vertical component. For the car on an 8% grade, for every 100 meters traveled horizontally, the elevation increases by 8 meters. This slope increases the amount of work needed, as the car has to fight gravity in addition to overcoming friction and wind resistance.

To solve for the power needed on an incline, we use the formula for work against gravity:

• \( Work = m \cdot g \cdot \frac{8}{100} \cdot d \)

where \( m \) is the mass of the car, \( g \) is gravitational acceleration (9.8 m/s²), and \( d \) is the distance traveled.

Given the constant speed, the total required power to move uphill combines both the power needed to overcome gravity and the continuous power against friction and wind:

• \( Power_{total} = Power_{friction} + Power_{gravity} \)

Constant Speed

Achieving and maintaining a constant speed is essential in understanding forces and energy in motion. At constant speed, all opposing forces such as friction and wind resistance are balanced by the power of the engine.

For the car moving on a level surface, this balance requires putting out 16 kW to simply offset these forces while no additional power is needed for acceleration.

When moving uphill, additional power is required not to accelerate but to counteract gravity. Similarly, when coasting downhill, gravity works to the vehicle's advantage, meaning that the engine does not need to provide additional power beyond overcoming friction and wind resistance. In fact, on a steep enough downgrade, the car's own weight can handle these forces and allow it to coast at the desired speed without engine power.

This balance of forces is crucial not only in vehicles but in many real-world applications of Newtonian mechanics, emphasizing why power and work calculations are fundamental in physics.