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When it is burned, 1 gallon of gasoline produces \(1.3 \times 10^{8} \mathrm{J}\) of energy. A \(1500-\mathrm{kg}\) car accelerates from rest to 37 \(\mathrm{m} / \mathrm{s}\) in 10 \(\mathrm{s}\) . The engine of this car is only 15\(\%\) efficient (which is typical), meaning that only 15\(\%\) of the energy from the combustion of the gasoline is used to accelerate the car. The rest goes into things like the internal kinetic energy of the engine parts as well as heating of the exhaust air and engine. (a) How many gallons of gasoline does this car use during the acceleration? (b) How many such accelerations will it take to burn up 1 gallon of gas?

Short Answer

Expert verified
(a) The car uses approximately 0.0527 gallons. (b) It takes about 19 accelerations to burn 1 gallon.

Step by step solution

01

Calculate the car's kinetic energy (KE)

We first need to calculate the kinetic energy of the car when it reaches the speed of 37 m/s. Use the kinetic energy formula: \( KE = \frac{1}{2}mv^2 \). Substitute \( m = 1500\, \text{kg} \) and \( v = 37\, \text{m/s} \):\[ KE = \frac{1}{2} \times 1500 \times 37^2 = 1,027,500 \, \text{J} \].
02

Calculate the useful energy from the gasoline

With the engine efficiency given as 15\%, the useful energy produced by burning the gasoline to achieve this kinetic energy will be: \( \text{Useful Energy} = \frac{KE}{\text{Efficiency}} = \frac{1,027,500}{0.15} = 6,850,000 \, \text{J} \).
03

Determine gallons of gasoline used for acceleration

The energy released by burning 1 gallon of gasoline is \( 1.3 \times 10^8 \, \text{J} \). To find out how many gallons are used, we solve: \[ \text{Gallons Used} = \frac{\text{Useful Energy Needed}}{\text{Energy per Gallon}} = \frac{6,850,000}{1.3 \times 10^8} \approx 0.0527 \, \text{gallons} \].
04

Calculate number of accelerations per gallon

To find the number of times the car can accelerate using 1 gallon of gasoline, calculate: \[ \text{Number of Accelerations} = \frac{1}{\text{Gallons Used per Acceleration}} = \frac{1}{0.0527} \approx 18.98 \]. Therefore, the car can accelerate approximately 19 times with 1 gallon of gasoline.

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Key Concepts

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

Kinetic Energy Calculation
Understanding the concept of kinetic energy is pivotal in grasping how vehicles move. Kinetic energy is the energy a body possesses due to its motion. For a car accelerating, this energy depends on both its mass and velocity. The formula to calculate kinetic energy is given by: \[ KE = \frac{1}{2}mv^2 \]where \( m \) is the mass of the object and \( v \) is its velocity. - In our example, the car has a mass of 1500 kg and accelerates to a velocity of 37 m/s. - By substituting these values into the formula, we can calculate the car's kinetic energy as 1,027,500 Joules.This energy tells us how much work was required to get the car from a standstill to its operating speed. Being familiar with these calculations not only helps in understanding energy usage and efficiency but also plays a crucial role in designing more energy-efficient automobiles.
Fuel Efficiency
Fuel efficiency is a measure of how effectively a vehicle converts the chemical energy stored in fuel into useful work for motion. This efficiency is often hindered by energy losses in various components of the car, such as:
  • Internal friction in the engine and moving parts
  • Thermal losses through exhaust and engine heating
For our scenario, the car engine is only 15% efficient. This means only 15% of the fuel's energy is used to accelerate the car, with the remaining energy lost. - The useful energy needed for acceleration is calculated by dividing the kinetic energy by the efficiency. - For this car, achieving a kinetic energy of 1,027,500 Joules requires 6,850,000 Joules of input energy from the gasoline. Understanding fuel efficiency aids in making decisions that are environmentally conscious and economically beneficial, encouraging practices that reduce fuel consumption and emissions.
Physics of Motion
The physics of motion encompasses the principles governing how objects move. In terms of vehicle dynamics, several key physics principles come into play: - **Newton's Laws of Motion:** These laws describe how forces interact with objects to influence motion, such as how a car moves when the engine exerts force on the wheels. - **Acceleration and Velocity:** These terms describe changes in motion. Acceleration refers to the rate of change of velocity that, in this exercise, brings the car from rest to a speed of 37 m/s in 10 seconds. Each of these principles contributes to either the increase or decrease in the vehicle's momentum, thereby affecting its acceleration and deceleration phases. Understanding these dynamics helps one appreciate the complexity involved in designing automobiles that respond predictably and effectively to the forces acting on them.

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Most popular questions from this chapter

An empty crate is given an initial push down a ramp, starting it with a speed \(v_{0},\) and reaches the bottom with speed \(v\) and kinetic energy \(K .\) Some books are now placed in the crate, so that the total mass is quadrupled. The coefficient of kinetic friction is constant and air resistance is negligible. Starting again with \(v_{0}\) at the top of the ramp, what are the speed and kinetic energy at the bottom? Explain the reasoning behind your answers.

A spring stores potential energy \(U_{0}\) when it is compressed a distance \(x_{0}\) from its uncompressed length. (a) In terms of \(U_{0},\) how much energy does it store when it is compressed (i) twice as much and (ii) half as much? (b) In terms of \(x_{0},\) how much must it be compressed from its uncompressed length to store (i) twice as much energy and (ii) half as much energy?

A truck with mass \(m\) has a brake failure while going down an ?cy mountain road of constant downward slope angle \(\alpha\) (Fig. 7.40\()\) Initially the truck is moving downhill at speed \(v_{0}\) - After careening downhill a distance \(L\) with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle \(\boldsymbol{\beta}\) . The truck rump has a soft sand suffice for which the coefficient of rolling friction is \(\mu_{r}\) What is the distance that the truck moves up the rump before coming to a halt? Solve using energy methods.

You are designing a delivery ramp for crates containing exercise equipment. The \(1470-\mathrm{N}\) crates will move at 1.8 \(\mathrm{m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a \(550-\mathrm{N}\) kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 8.0 \(\mathrm{m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the force constant of the spring that will be needed in order to meet the design criteria.

A wooden block with mass 1.50 \(\mathrm{kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A\) ). When the spring is released, it projects the block up the incline. At point \(B,\) a distance of 6.00 \(\mathrm{mup}\) the incline from \(A\) , the block is moving up the incline at 7.00 \(\mathrm{m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{\mathrm{k}}=0.50 .\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

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