Ch 09: Work and Kinetic Energy

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 09: Work and Kinetic Energy

Problem 63a

Chapter 9, Problem 63a

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. What is the maximum acceleration of the Porsche on a concrete surface where μ_s = 1.00? Hint: What force pushes the car forward?

Verified step by step guidance

Step 1: Understand the problem. The goal is to calculate the maximum acceleration of the Porsche on a concrete surface. To do this, we need to determine the force that pushes the car forward, which is limited by the static friction between the drive wheels and the surface. The static friction force is given by \( F_{\text{friction}} = \mu_s \cdot F_{\text{normal}} \), where \( \mu_s \) is the coefficient of static friction and \( F_{\text{normal}} \) is the normal force acting on the drive wheels.

Step 2: Calculate the normal force on the drive wheels. The total weight of the car and driver is \( F_{\text{gravity}} = m \cdot g \), where \( m = 1480 \; \text{kg} \) and \( g = 9.8 \; \text{m/s}^2 \). Since two-thirds of the weight is over the drive wheels, the normal force on the drive wheels is \( F_{\text{normal}} = \frac{2}{3} \cdot F_{\text{gravity}} \).

Step 3: Determine the maximum static friction force. Using the formula \( F_{\text{friction}} = \mu_s \cdot F_{\text{normal}} \), substitute \( \mu_s = 1.00 \) and the value of \( F_{\text{normal}} \) calculated in Step 2 to find the maximum force that can push the car forward.

Step 4: Relate the friction force to acceleration. The maximum acceleration is determined by Newton's second law, \( F = m \cdot a \). Rearrange this equation to solve for acceleration: \( a = \frac{F_{\text{friction}}}{m} \). Substitute the value of \( F_{\text{friction}} \) from Step 3 and the mass \( m = 1480 \; \text{kg} \) to find the maximum acceleration.

Step 5: Verify the assumptions. Ensure that the engine power and drivetrain losses are not directly relevant to the calculation of maximum acceleration, as the problem focuses on the force due to static friction. The hint emphasizes that the force pushing the car forward is limited by the frictional force, not the engine power.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Power and Efficiency

Power is the rate at which work is done or energy is transferred. In this context, the engine's rated power of 217 hp indicates its maximum output. However, efficiency plays a crucial role, as only 70% of this power is effectively used to propel the car forward after accounting for losses in the engine and drivetrain. Understanding how to calculate the usable power is essential for determining the car's performance.

Friction and Traction

Friction is the force that resists the relative motion of solid surfaces, and it is crucial for vehicle acceleration. The coefficient of static friction (μₛ) quantifies the maximum frictional force before sliding occurs. In this case, with μₛ = 1.00 on a concrete surface, the frictional force provides the necessary traction for the car to accelerate. Calculating this force is vital for determining the car's maximum acceleration.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). This principle is fundamental in analyzing the car's motion. By applying this law, one can relate the net force generated by the frictional force to the car's mass to find the maximum acceleration achievable under the given conditions.

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Related Practice

Textbook Question

When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of the air. Model a cyclist as having cross-section area 0.45 m² and, because the human body is not aerodynamically shaped, a drag coefficient of 0.90. Use 1.2 kg/m³ as the density of air at room temperature. The food calorie is equivalent to 4190 J. How many calories does the cyclist burn if he rides over level ground at 7.3 m/s for 1 h?

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Textbook Question

When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of the air. Model a cyclist as having cross-section area 0.45 m² and, because the human body is not aerodynamically shaped, a drag coefficient of 0.90. Use 1.2 kg/m³ as the density of air at room temperature. Metabolic power is the rate at which your body 'burns' fuel to power your activities. For many activities, your body is roughly 25% efficient at converting the chemical energy of food into mechanical energy. What is the cyclist's metabolic power while cycling at 7.3 m/s?

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Textbook Question

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. How long does it take the Porsche to reach the maximum power output?

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Textbook Question

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. If the Porsche accelerates at a_max, what is its speed when it reaches maximum power output?

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Textbook Question

A hydroelectric power plant uses spinning turbines to transform the kinetic energy of moving water into electric energy with 80% efficiency. That is, 80% of the kinetic energy becomes electric energy. A small hydroelectric plant at the base of a dam generates 50 MW of electric power when the falling water has a speed of 18 m/s. What is the water flow rate - kilograms of water per second - through the turbines?

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Textbook Question

A farmer uses a tractor to pull a 150 kg bale of hay up a 15° incline to the barn at a steady 5.0 km/h. The coefficient of kinetic friction between the bale and the ramp is 0.45. What is the tractor's power output?

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