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Chapter 6: Problem 57

Interactive Solution \(6.57\) at offers a model for solving this problem. 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) 32.8 kW; (b) 51.0 kW.

Step by step solution

01

Calculating Acceleration

To start, we need to calculate the acceleration of the car using the formula for acceleration: \( a = \frac{\Delta v}{\Delta t} \). In this problem, \( \Delta v = 20.0 \, \text{m/s} \) and \( \Delta t = 5.6 \, \text{s} \). Plug these values into the formula: \( a = \frac{20.0}{5.6} = 3.57 \, \text{m/s}^2 \).
02

Calculating Mass of the Car

Next, use the weight of the car to calculate its mass. The weight (\( W \)) is the force of gravity acting on the car, given by \( W = mg \), where \( m \) is the mass and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. For part (a), \( W = 9.0 \times 10^3 \, \text{N} \). So, \( m = \frac{9.0 \times 10^3}{9.8} = 918.37 \, \text{kg} \).
03

Calculating Work Done

Calculate the work done using the change in kinetic energy. The work done on the car is given by \( \Delta K = \frac{1}{2} m v^2 \), since the car starts from rest. For part (a), substitute the values: \( \Delta K = \frac{1}{2} \times 918.37 \times (20)^2 = 183674 \, \text{J} \).
04

Calculating Power

Finally, power is the work done per unit time. Use the formula \( P = \frac{\Delta K}{\Delta t} \). Substitute \( \Delta K = 183674 \, \text{J} \) and \( \Delta t = 5.6 \, \text{s} \) for part (a): \( P = \frac{183674}{5.6} = 32799 \, \text{W} \) or \( 32.8 \, \text{kW} \).
05

Repeat for Part (b)

Follow the same steps for part (b) with the new weight \( W = 1.4 \times 10^4 \, \text{N} \). Calculate mass: \( m = \frac{1.4 \times 10^4}{9.8} = 1428.57 \, \text{kg} \). Calculate work done: \( \Delta K = \frac{1}{2} \times 1428.57 \times (20)^2 = 285714.29 \, \text{J} \). Calculate power: \( P = \frac{285714.29}{5.6} = 51020 \, \text{W} \) or \( 51.0 \, \text{kW} \).

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Key Concepts

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

Uniform Acceleration
Uniform acceleration occurs when the velocity of an object changes at a constant rate. In our exercise, the car experiences uniform acceleration as it speeds up from rest to 20 m/s in 5.6 seconds. This means that the car's velocity increases by the same amount every second.

To find the acceleration, we use the formula: \[a = \frac{\Delta v}{\Delta t}\]where \(\Delta v\) is the change in velocity, and \(\Delta t\) is the time period over which the change occurs. In this case, the velocity changes by 20 m/s in 5.6 seconds:
- \(a = \frac{20.0}{5.6} = 3.57 \, \text{m/s}^2\)
- This value represents how quickly the car is picking up speed.
Kinetic Energy
Kinetic energy is the energy that an object has due to its motion. It is an important concept in vehicle mechanics, as moving objects need kinetic energy to keep going.

The kinetic energy \(K\) of an object can be calculated using:\[K = \frac{1}{2} mv^2\]where \(m\) is the mass of the object, and \(v\) is its velocity. For the car, we calculated kinetic energy because it helps in determining the total work done to accelerate the car.
- For a car of mass 918.37 kg reaching 20 m/s, the kinetic energy change \(\Delta K\) was:\[\Delta K = \frac{1}{2} \times 918.37 \times (20)^2 = 183674 \, \text{J} \]
Power Calculation
Power in the context of vehicle mechanics is the rate at which work is done or energy is transferred. When a car accelerates, it requires power to meet the energy demands of changing speed.

The formula to calculate power \(P\) is:\[P = \frac{W}{t}\]where \(W\) is the work done or energy transferred in joules, and \(t\) is the time in seconds over which the work is done. The work in this scenario is the change in kinetic energy:
  • For a 9,000 N car, the power needed was \(P = \frac{183674}{5.6} = 32799 \, \text{W}\) or 32.8 kW.
  • For a 14,000 N car, it required \(P = \frac{285714.29}{5.6} = 51020 \, \text{W}\) or 51.0 kW.

This power calculation is crucial in understanding how engines need to be designed to provide sufficient acceleration.
Newton's Second Law
Newton's Second Law of Motion is central to understanding how forces affect the movement of a car. It states that the force acting on an object is equal to the mass of the object times its acceleration:\[F = ma\]This principle explains why heavier vehicles require more force to achieve the same acceleration compared to lighter ones.
- In our example, once the car's mass is known, Newton's Second Law helps in determining how much force is necessary to reach a certain acceleration.
  • For a car with a mass of 918.37 kg and acceleration of 3.57 m/s虏, the force is: \(F = 918.37 \times 3.57\)
  • This also underscores how a car's engine needs to generate sufficient power to overcome its inertial resistance and achieve movement.
Newton's Second Law is a foundational concept for understanding vehicle dynamics and the forces involved in motion.

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Most popular questions from this chapter

A basketball of mass \(0.60 \mathrm{~kg}\) is dropped from rest from a height of \(1.05 \mathrm{~m}\). It rebounds to a height of \(0.57 \mathrm{~m}\) (a) How much mechanical energy was lost during the collision with the floor? (b) A basketball player dribbles the ball from a height of \(1.05 \mathrm{~m}\) by exerting a constant downward force on it for a distance of \(0.080 \mathrm{~m}\). In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.05 \(\mathrm{m},\) what is the magnitude of the force?

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.

The brakes of a truck cause it to slow down by applying a retarding force of \(3.0 \times 10^{3} \mathrm{~N}\) to the truck over a distance of \(850 \mathrm{~m}\). What is the work done by this force on the truck? Is the work positive or negative? Why?

A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest of the second hill is circular, with a radius of \(r=36 \mathrm{~m} .\) Neglect friction and air resistance. What must be the height \(h\) of the first hill so that the skier just loses contact with the snow at the crest of the second hill?

A pitcher throws a \(0.140-\mathrm{kg}\) baseball, and it approaches the bat at a speed of \(40.0 \mathrm{~m} / \mathrm{s}\). The bat does \(W_{n c}=70.0 \mathrm{~J}\) of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is \(25.0 \mathrm{~m}\) above the point of impact.
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