/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 (a) How many joules of kinetic e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 10

(a) How many joules of kinetic energy does a \(750-\mathrm{kg}\) auto- mobile traveling at a typical highway speed of 65 \(\mathrm{mi} / \mathrm{h}\) have? (b) By what factor would its kinetic energy decrease if the car traveled half as fast? (c) How fast (in mi/h) would the car have to travel to have half as much kinetic energy as in part (a)?

Short Answer

Expert verified
(a) 317307 J; (b) Kinetic energy decreases by a factor of 4; (c) 45.97 mi/h.

Step by step solution

01

Convert Units

First, convert the speed from miles per hour to meters per second, since the mass is given in kilograms and we will calculate energy in joules. 1 mile = 1609.34 meters, 1 hour = 3600 seconds. So, 65 mi/h = 65 * 1609.34 m / 3600 s = approximately 29.06 m/s.
02

Calculate Initial Kinetic Energy

The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \).Substitute the values: \( m = 750 \text{ kg} \) and \( v = 29.06 \text{ m/s} \).\[ KE = \frac{1}{2} \times 750 \times (29.06)^2 \approx 317307 \text{ J} \]
03

Consider Half Speed Scenario

If the speed is halved, the new speed \( v_{new} = \frac{29.06}{2} \approx 14.53 \text{ m/s} \).Substitute into the kinetic energy formula: \( KE_{new} = \frac{1}{2} \times 750 \times (14.53)^2 \).\[ KE_{new} \approx 79326.75 \text{ J} \]
04

Calculate Factor Reduction for Half Speed

To find the factor by which the kinetic energy decreases, compare \( KE \) with \( KE_{new} \).The factor is \( \frac{79326.75}{317307} \approx 0.25 \).This shows that the kinetic energy is reduced by a factor of 4 when the speed is halved.
05

Determine Speed for Half Kinetic Energy in Part a

To have half as much kinetic energy as in part (a), set \( KE_{half} = \frac{317307}{2} \approx 158653.5 \text{ J} \).Use the kinetic energy formula \( KE = \frac{1}{2}mv^2 \) to find \( v \).\[ 158653.5 = \frac{1}{2} \times 750 \times v^2 \]Solve for \( v \): \[ v = \sqrt{\frac{2 \times 158653.5}{750}} \approx 20.54 \text{ m/s} \]
06

Convert Back to Miles Per Hour

Convert \( v = 20.54 \text{ m/s} \) back to miles per hour:\( 20.54 \text{ m/s} \times \frac{3600 \text{ s}}{1609.34 \text{ m}} = \approx 45.97 \text{ mi/h} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Physics Problem Solving
Physics problem solving involves breaking down complex scenarios into manageable steps. Here, we tackled a problem involving kinetic energy by using a systematic approach:
  • Understand what the problem is asking. Break it into smaller parts: Calculate kinetic energy, consider changes in speed, and find equivalent energy states.
  • Identify the formulas and physical laws relevant to the problem. For kinetic energy, we need the formula \( KE = \frac{1}{2}mv^2 \).
  • Use logical reasoning to adapt known principles to the specific question. Determine how changes in velocity alter kinetic energy.
Each step relies on a clear grasp of the physics concepts involved and ensures that the correct method is applied for problem-solving success.
Unit Conversion
Unit conversion is crucial in physics as it ensures consistency in calculations. In this problem, the speed was given in miles per hour, whereas calculations of energy in joules require meters per second. Here's how we handled it:
  • Recognize the need for conversion: When units don't match, conversion is essential.
  • Use known equivalents for conversion: 1 mile = 1609.34 meters, 1 hour = 3600 seconds.
  • Perform the conversion: \( 65 \text{ mi/h} = 65 \times \frac{1609.34 \text{ m}}{3600 \text{ s}} \approx 29.06 \text{ m/s} \).
By converting units, we ensure that our calculations remain accurate and consistent with the physical quantities involved.
Speed and Velocity
Speed and velocity are terms that describe motion, but physics often requires precise use of these terms for calculation.
  • Speed is a scalar quantity 鈥� it only has magnitude.
  • Velocity is a vector 鈥� it includes both magnitude and direction.
In energy calculations, we often refer to speed because we're concerned only with how fast an object is moving regardless of direction. Changes in speed have significant effects on kinetic energy, due to its square relationship in the formula. Halving the speed to calculate kinetic energy shows that \( v^2 \) leads to only a quarter of the original energy.
Energy Calculations
Energy calculations often involve determining energies at different states to understand their relations. Using the kinetic energy formula, we calculated the energy of a moving car.
  • Kinetic energy formula: \( KE = \frac{1}{2}mv^2 \).
  • Calculate initial energy: For the given car, \( KE = \frac{1}{2} \times 750 \times (29.06)^2 \approx 317307 \text{ J} \).
  • Understand effect of speed reduction: Halved speed results in one-quarter of the original energy.
Moreover, calculating energy for half speed and determining speed for half of initial energy helps illustrate the direct impact speed has on kinetic energy. Energy calculations require careful substitution of values and understanding of proportional changes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A tow truck pulls a car 5.00 \(\mathrm{km}\) along a horizontal roadway using a cable having a tension of 850 \(\mathrm{N}\) . (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part \((a) ?(c)\) How much work does gravity do on the car in part (a)?

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).

You are a member of an Alpine Rescue Team. You must project a box of supphes up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_{\mathbf{k}}\) . Use the work-energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g, h, \mu_{k},\) and \(\alpha .\)

Pushing a Cat. Your cat "Ms" (mass 7.00 \(\mathrm{kg}\) ) is trying to make it to the top of a frictionless ramp 2.00 \(\mathrm{m}\) long and inclined upward at \(30.0^{\circ}\) above the horizontal. Since the poor cat can't get any traction on the ramp, you push her up the entire length of the ramp by excring a constant \(100-\mathrm{N}\) force parallel to the ramp. If Ms. takes a running start so that she is moving at 2.40 \(\mathrm{m} / \mathrm{s}\) at the bottom of the ramp, what is her speed when she reaches the top of the incline? Use the work-energy theorem.

Rescue. Your friend (mass 65.0 \(\mathrm{kg}\) ) is standing on the ice in the middle of a frozen pond. There is very hittle friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 \(\mathrm{s}\) and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) while you remain at rest. What is the average power supplied by the force you applied?
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Particle Model of Matter

Read Explanation

Modern Physics

Read Explanation

Medical Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Translational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.