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A car accelerates uniformly from rest to \(20.0 \mathrm{~m} / \mathrm{s}\) in \(5.6 \mathrm{~s}\) along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is \(9.0 \times 10^{3} \mathrm{~N},\) and \((\mathrm{b})\) the weight of the car is \(1.4 \times 10^{4} \mathrm{~N}\)

Short Answer

Expert verified
(a) 32870.46 W (b) 51020.76 W

Step by step solution

01

Determine Mass of the Car

The weight of the car is given by the formula \( W = m \cdot g \) where \( W \) is the weight, \( m \) is the mass, and \( g \) is the acceleration due to gravity (\( 9.8 \ \mathrm{m/s^2} \)). For part (a), the weight is \( 9.0 \times 10^3 \ \mathrm{N} \). Calculate \( m = \frac{W}{g} = \frac{9.0 \times 10^3}{9.8} \approx 918.37 \ \mathrm{kg} \). For part (b), the weight is \( 1.4 \times 10^4 \ \mathrm{N} \). Calculate \( m = \frac{1.4 \times 10^4}{9.8} \approx 1428.57 \ \mathrm{kg} \).
02

Calculate Final Kinetic Energy

The kinetic energy \( K \) of an object is given by the formula \( K = \frac{1}{2}mv^2 \). For part (a), using \( m = 918.37 \ \mathrm{kg} \) and \( v = 20 \ \mathrm{m/s} \), the kinetic energy is \( K = \frac{1}{2} \times 918.37 \times 20^2 = 183674.57 \ \mathrm{J} \). For part (b), using \( m = 1428.57 \ \mathrm{kg} \), \( K = \frac{1}{2} \times 1428.57 \times 20^2 = 285714.29 \ \mathrm{J} \).
03

Calculate Average Power

Average power \( P \) is calculated by dividing the work done or change in kinetic energy by the time taken. So, \( P = \frac{K}{t} \). For part (a), \( P = \frac{183674.57}{5.6} \approx 32870.46 \ \mathrm{W} \). For part (b), \( P = \frac{285714.29}{5.6} \approx 51020.76 \ \mathrm{W} \).

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Key Concepts

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

Uniform Acceleration
Uniform acceleration is a concept in physics that occurs when an object's velocity changes at a constant rate. This means that the object is speeding up or slowing down uniformly over time. In the exercise, the car accelerates uniformly from rest, which allows us to use straightforward calculations. Here is how uniform acceleration can be understood better:

The key characteristic of uniform acceleration is that the acceleration value remains the same throughout the motion. This makes calculations more predictable and easier to perform.

For example, if a car starts from rest and reaches a speed of 20 m/s in 5.6 seconds, as in the exercise, it has a constant acceleration. This means each second it speeds up by a consistent amount until reaching its final speed.

Understanding uniform acceleration is important for accurately predicting the behavior of objects in motion, especially when considering additional forces like gravity or friction in more complex scenarios. Always remember that in uniform acceleration:
  • Initial velocity and time are crucial for calculating acceleration.
  • Acceleration enables us to predict the future velocity and position of an object.
These basic concepts provide a foundation for more complex motion studies, helping students grasp fundamental principles of dynamics.
Kinetic Energy Calculation
Kinetic energy represents the energy that an object possesses due to its motion. It is calculated using the formula:
\[ K = \frac{1}{2}mv^2 \]
where \( m \) is mass and \( v \) is velocity.

In the context of the exercise, calculating kinetic energy helps us in determining the energy involved in moving the car from rest to a certain velocity. Here's a further explanation:
  • Kinetic energy depends on both mass and the square of velocity. This means that even a small increase in velocity can lead to a large increase in kinetic energy.
  • The calculation of kinetic energy is pivotal because it facilitates understanding of how quickly an object can accelerate and the amount of energy it uses when it moves.
  • In problems involving changes in motion, kinetic energy is often a starting point for calculating other physical quantities, such as work done or power needed, as seen in the exercise.
Grasping the basics of kinetic energy is essential for diving deeper into other related topics like potential energy, mechanical energy, and conservation of energy principles. It provides a critical link between an object’s speed and its energy needs.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that links the work done on an object to its change in kinetic energy. Essentially, it states that the total work done by forces on an object equals its change in kinetic energy. This principle is crucial for solving many physics problems, such as the one in this exercise.

In the uniform acceleration problem, knowing the work-energy principle helps us calculate how much energy was required to bring the car to its final velocity.

Here is a further breakdown:
  • Work is defined as the force applied to an object times the distance the object moves in the direction of that force.
  • The work-energy principle can be expressed as:
\[W = \Delta K\]
This tells us that the work done on the car is equal to its change in kinetic energy from rest to its final velocity of 20 m/s.

This powerful relationship makes it easier to solve for average power, as power is simply the rate at which work is done over time:
  • Average power is calculated by dividing the work (or change in kinetic energy) by time.
  • It provides insights into how efficient a system converts energy to motion.
Utilizing the work-energy principle leads to a deeper understanding of energy transfer, system efficiency, and the forces at play in dynamic scenarios.

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Most popular questions from this chapter

A \(55.0-\mathrm{kg}\) skateboarder starts out with a speed of \(1.80 \mathrm{~m} / \mathrm{s} .\) He does \(+80.0 \mathrm{~J}\) of work on himself by pushing with his feet against the ground. In addition, friction does \(-265 \mathrm{~J}\) of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is \(6.00 \mathrm{~m} / \mathrm{s}\). (a) Calculate the change \(\left(\Delta \mathrm{PE}=\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the gravitational potential energy. (b) How much has the vertical height of the skater changed, and is the skater above or below the starting point?

A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled's displacement, which is along the \(+x\) axis. As a result, the kinetic energy of the sled increases by \(38 \% .\) By what percentage would the sled's kinetic energy have increased if this force had pointed \(62^{\circ}\) above the \(+x\) axis?

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. (a) Does the kinetic frictional force do positive, negative, or zero work? Provide a reason for your answer. (b) Does the total mechanical energy of the student increase, decrease, or remain the same as she descends the slide? Why? (c) If the kinetic frictional force does work, how is this work related to the change in the total mechanical energy of the student? The student has a mass of \(83.0 \mathrm{~kg}\) and the height of the water slide is \(11.8 \mathrm{~m}\). If the kinetic frictional force does \(-6.50 \times 10^{3} \mathrm{~J}\) of work, how fast is the student going at the bottom of the slide?

A husband and wife take turns pulling their child in a wagon along a horizontal sidewalk. Each exerts a constant force and pulls the wagon through the same displacement. They do the same amount of work, but the husband's pulling force is directed \(58^{\circ}\) above the horizontal, and the wife's pulling force is directed \(38^{\circ}\) above the horizontal. The husband pulls with a force whose magnitude is \(67 \mathrm{~N}\). What is the magnitude of the pulling force exerted by his wife?

The hammer throw is a track-and-field event in which a \(7.3-\mathrm{kg}\) ball (the "hammer'), starting from rest, is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion. In one throw, the hammer is given a speed of \(29 \mathrm{~m} / \mathrm{s}\). For comparison, a .22 caliber bullet has a mass of \(2.6 \mathrm{~g}\) and, starting from rest, exits the barrel of a gun with a speed of \(410 \mathrm{~m} / \mathrm{s}\). Determine the work done to launch the motion of (a) the hammer and (b) the bullet.

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