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Chapter 6: Problem 100

Automotive Power I. A truck engine transmits 28.0 \(\mathrm{kW}(37.5 \mathrm{hp})\) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0 \(\mathrm{km} / \mathrm{h}\) (37.3 \(\mathrm{mi} / \mathrm{h} )\) on a level road. (a) What is the resisting force acting on the truck? (b) Assume that 65\(\%\) of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at 30.0 \(\mathrm{km} / \mathrm{h} ?\) At 120.0 \(\mathrm{km} / \mathrm{h} ?\) Give your answers in kilowatts and in horsepower.

Short Answer

Expert verified
(a) 1680 N; (b) 10.32 kW (13.84 hp) at 30 km/h, 114.80 kW (153.90 hp) at 120 km/h.

Step by step solution

01

Determine Resisting Force

The power transmitted to the wheels is given as \( P = 28.0 \, \text{kW} \). The truck is moving at a constant velocity of \( v = 60.0 \, \text{km/h} = 16.67 \, \text{m/s} \). To find the resisting force \( F \), use the formula \( P = F \times v \). Rearrange it to \( F = \frac{P}{v} \). Thus, \( F = \frac{28000 \, \text{W}}{16.67 \, \text{m/s}} \approx 1680.02 \, \text{N} \).
02

Calculate Individual Resistances

Assume 65% of the resisting force is due to rolling friction, \( F_{rolling} = 0.65 \times 1680.02 \, \text{N} = 1092.01 \, \text{N} \), and 35% is due to air resistance, \( F_{air} = 0.35 \times 1680.02 \, \text{N} = 588.01 \, \text{N} \).
03

Calculate Force at Different Speeds

The force of rolling friction \( F_{rolling} = 1092.01 \, \text{N} \) is constant. The air resistance scales with the square of the speed. Thus, - At 30.0 km/h (8.33 m/s), \( F_{air,30} = 588.01 \, \text{N} \times \left( \frac{8.33}{16.67} \right)^2 \approx 147.00 \, \text{N} \). - At 120.0 km/h (33.33 m/s), \( F_{air,120} = 588.01 \, \text{N} \times \left( \frac{33.33}{16.67} \right)^2 \approx 2352.04 \, \text{N} \).
04

Calculate Required Power at Different Speeds

Calculate the total force at each speed: - At 30 km/h: \( F_{total,30} = F_{rolling} + F_{air,30} \approx 1092.01 \, \text{N} + 147.00 \, \text{N} = 1239.01 \, \text{N} \). - At 120 km/h: \( F_{total,120} = F_{rolling} + F_{air,120} \approx 1092.01 \, \text{N} + 2352.04 \, \text{N} = 3444.05 \, \text{N} \).
05

Calculate Power at Each Speed

Use the power formula for each total force:- At 30 km/h: \( P_{30} = F_{total,30} \times v_{30} = 1239.01 \, \text{N} \times 8.33 \, \text{m/s} \approx 10320.69 \, \text{W} = 10.32 \, \text{kW} \) or \( 13.84 \, \text{hp} \).- At 120 km/h: \( P_{120} = F_{total,120} \times v_{120} = 3444.05 \, \text{N} \times 33.33 \, \text{m/s} \approx 114798.62 \, \text{W} = 114.80 \, \text{kW} \) or \( 153.90 \, \text{hp} \).

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Key Concepts

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

Resisting Force
When a vehicle is moving at a constant velocity, meaning its speed and direction aren't changing, the power generated by the engine is balanced by the resisting forces acting against the vehicle. These resisting forces can include rolling friction and air resistance. The resisting force can be calculated using the relationship between power (P) and velocity (v). The equation is expressed as \( P = F \times v \), where \(F\) is the resisting force. By rearranging this formula to \( F = \frac{P}{v} \), we can find the total resisting force acting on the vehicle. In this example, the truck with a power output of 28.0 kW moving at 60.0 km/h has a calculated resisting force of approximately 1680 N. This force represents the sum total of all forces acting to slow down the truck.
Rolling Friction
Rolling friction is a form of friction that occurs when an object rolls over a surface. Unlike sliding friction, which is often more significant, rolling friction is generally weaker and mostly depends on the nature of the surfaces in contact and the weight of the object. In this exercise, rolling friction is responsible for 65% of the truck's total resisting force. This can be calculated as \( F_{rolling} = 0.65 \times 1680 \, \text{N} = 1092.01 \, \text{N} \). Rolling friction is independent of speed, meaning it doesn't change regardless of how fast the truck is moving. It remains constant at 1092.01 N, whether the truck travels at 30 km/h or at 120 km/h.
Air Resistance
Air resistance, also known as drag, is a force that opposes the motion of an object through a fluid, which in this case is air. Unlike rolling friction, air resistance increases with the square of the speed of the vehicle. Therefore, it becomes a more significant factor as speed increases. In this scenario, air resistance makes up the remaining 35% of the truck's resisting force, calculated initially as \( F_{air} = 0.35 \times 1680 \, \text{N} = 588.01 \, \text{N} \). When the truck moves at different speeds, these values change: at 30 km/h, \( F_{air,30} \approx 147.00 \, \text{N} \), and at 120 km/h, \( F_{air,120} \approx 2352.04 \, \text{N} \). Thus, at higher speeds, air resistance becomes a dominant force that the engine must overcome, requiring substantially more power compared to lower speeds.

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Most popular questions from this chapter

CP A 20.0 -kg crate sits at rest at the bottom of a 15.0 -m-long ramp that is inclined at \(34.0^{\circ}\) above the horizontal. A constant horizontal force of 290 \(\mathrm{N}\) is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 \(\mathrm{N}\) . (a) What is the total work done on the crate during its motion from the bottom to the top of the ramp? (b) How much time does it take the crate to travel to the top of the ramp?

CALC Rotating Bar. A thin, uniform 12.0 -kg bar that is 2.00 \(\mathrm{m}\) long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar? (Hint. Different points in the bar have different speeds. Break the bar up into infinitesimal segments of mass dm and integrate to add up the kinetic energies of all these segments.)

Six diesel units in series can provide 13.4 \(\mathrm{MW}\) of power to the lead car of a freight train. The diesel units have total mass \(1.10 \times 10^{6} \mathrm{kg}\) . The average car in the train has mass \(8.2 \times 10^{4} \mathrm{kg}\) and requires a horizontal pull of 2.8 \(\mathrm{kN}\) to move at a constant 27 \(\mathrm{m} / \mathrm{s}\) on level tracks. (a) How many cars can be in the train under these conditions? (b) This would leave no power for accelerating or climbing hills. Show that the extra force needed to accelerate the train is about the same for a \(0.10-\mathrm{m} / \mathrm{s}^{2}\) acceleration or a 1.0\(\%\) slope (slope angle \(\alpha=\arctan 0.010 )\) . (c) With the 1.0\(\%\) slope, show that an extra 2.9 \(\mathrm{MW}\) of power is needed to maintain the \(27-\mathrm{m} / \mathrm{s}\) speed of the diesel units. (d) With 2.9 \(\mathrm{MW}\) less power available, how many cars can the six diesel units pull up a 1.0\(\%\) slope at a constant 27 \(\mathrm{m} / \mathrm{s} ?\)

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 \(\mathrm{km} / \mathrm{h}\) ?

CALC An object is attracted toward the origin with a force given by \(F_{x}=-k / x^{2}\) . (Gravitational and electrical forces have this distance dependence.) (a) Calculate the work done by the force \(F_{x}\) when the object moves in the \(x\) -direction from \(x_{1}\) to \(x_{2}\) . If \(x_{2}>x_{1},\) is the work done by \(F_{x}\) positive or negative? (b) The only other force acting on the object is a force that you exert with your hand to move the object slowly from \(x_{1}\) to \(x_{2} .\) How much work do you do? If \(x_{2}>x_{1},\) is the work you do positive or negative? (c) Explain the similarities and differences between your answers to parts (a) and (b).
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