/*! 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 66 If a car of mass \(1200 \mathrm{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 66

If a car of mass \(1200 \mathrm{~kg}\) is moving along a highway at \(120 \mathrm{~km} / \mathrm{h}\) what is the car's kinetic energy as determined by someone standing alongside the highway?

Short Answer

Expert verified
The car's kinetic energy is approximately 666,500 J.

Step by step solution

01

Convert the speed from km/h to m/s

To find the kinetic energy, we first need the velocity in meters per second (m/s). The initial given speed is \(120\, \text{km/h}\). To convert it to meters per second, use the formula \(v = \frac{{\text{speed in km/h} \times 1000}}{{3600}}\).\[ v = \frac{{120 \times 1000}}{{3600}} = \frac{{120000}}{{3600}} = 33.33 \, \text{m/s} \]
02

Use the kinetic energy formula

The formula for kinetic energy \(E_k\) of an object is given by \(E_k = \frac{1}{2} mv^2 \) where \(m\) is the mass and \(v\) is the velocity.Substitute \(m = 1200\, \text{kg}\) and \(v = 33.33\, \text{m/s}\) into the equation:\[ E_k = \frac{1}{2} \times 1200 \times (33.33)^2 \]
03

Calculate the expression

Now perform the calculation:\[ E_k = 600 \times 33.33^2 = 600 \times 1110.8889 = 666533.34 \, \text{J} \]
04

Round the result

Kinetic energy is typically expressed with a reasonable number of significant figures.Thus, \(E_k \approx 666500 \, \text{J}\).

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.

Conversion of Units
When dealing with physics problems, one essential step is the conversion of units. This is because it ensures consistency in equations and helps in comparing results effectively. In many exercises, like calculating kinetic energy, the speed and mass need to be converted into compatible units.

In this specific exercise, the car's speed is initially given in kilometers per hour (km/h). However, for the kinetic energy calculation, it is easier to work with speed in meters per second (m/s). To convert km/h to m/s, multiply by 1000 to switch kilometers to meters, then divide the result by 3600 to convert hours to seconds. The formula looks like this:
  • Speed in m/s = (Speed in km/h 脳 1000) / 3600


By applying this formula to a speed of 120 km/h, the conversion leads us to a velocity of 33.33 m/s. Proper conversion like this is crucial because using incorrect units is a common source of error in physics problem-solving.
Physics Problem Solving
Solving physics problems involves a systematic approach to find unknown quantities based on given data. This method can be broken down into steps that help in finding solutions effectively.

For the kinetic energy problem, the process begins by identifying known parameters:
  • Mass of the car: 1200 kg
  • Initial speed: 120 km/h


After identifying what is given, the next step is choosing the correct formulas and conversions. Here, converting speed from km/h to m/s was necessary to use the standard kinetic energy formula. Breaking a problem into parts not only clears confusion but enhances accuracy.

After that, the kinetic energy formula is applied to calculate energy. Such an approach is generally applicable in physics:
  • Read the problem carefully.
  • Identify what you know and what you need to find out.
  • Convert units if necessary, and use relevant formulas.
  • Solve step-by-step.


This methodical problem-solving strategy is a fundamental skill to develop in physics.
Kinetic Energy Formula
Kinetic energy is the energy of motion, quantified by the formula \( E_k = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity. This formula shows us that kinetic energy depends directly on the mass of the object and the square of its velocity.

You notice from the formula that even a small increase in speed can lead to a large increase in kinetic energy because velocity is squared.

In this exercise, we calculated the kinetic energy of a car by substituting its mass 1200 kg and a velocity of 33.33 m/s into the formula:
  • \( E_k = \frac{1}{2} \times 1200 \times (33.33)^2 \)


Calculating further led to \( E_k \approx 666,500 \) joules (rounded). This shows how the formula converts mass and speed into energy, allowing us to understand the dynamics of moving objects.

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 ice block floating in a river is pushed through a displacement \(\vec{d}=(15 \mathrm{~m}) \mathrm{i}-(12 \mathrm{~m}) \mathrm{j}\) along a straight embankment by rushing water, which exerts a force \(\vec{F}=(210 \mathrm{~N}) \mathrm{i}-(150 \mathrm{~N}) \mathrm{j}\) on the block. How much work does the force do on the block during the displacement?

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of \(12^{\circ}\) with the horizontal. The rope moves parallel to the slope with a constant speed of \(1.0 \mathrm{~m} / \mathrm{s}\). The force of the rope does \(900 \mathrm{~J}\) of work on the skier as the skier moves a distance of \(8.0 \mathrm{~m}\) up the incline. (a) If the rope moved with a constant speed of \(2.0 \mathrm{~m} / \mathrm{s}\), how much work would the force of the rope do on the skier as the skier moved a distance of \(8.0 \mathrm{~m}\) up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) \(1.0 \mathrm{~m} / \mathrm{s}\) and (c) \(2.0 \mathrm{~m} / \mathrm{s} ?\)

A single force acts on a \(3.0 \mathrm{~kg}\) particle-like object whose position is given by \(x=3.0 t-4.0 t^{2}+1.0 t^{3},\) with \(x\) in meters and \(t\) in seconds. Find the work done by the force from \(t=0\) to \(t=4.0 \mathrm{~s}\).

A CD case slides along a floor in the positive direction of an \(x\) axis while an applied force \(\vec{F}_{a}\) acts on the case. The force is directed along the \(x\) axis and has the \(x\) component \(F_{a x}=9 x-3 x^{2}\) with \(x\) in meters and \(F_{a x}\) in newtons. The case starts at rest at the position \(x=0,\) and it moves until it is again at rest. (a) Plot the work \(\vec{F}_{a}\) does on the case as a function of \(x\). (b) At what position is the work maximum, and (c) what is that maximum value? (d) At what position has the work decreased to zero? (e) At what position is the case again at rest?

A \(250 \mathrm{~g}\) block is dropped onto a relaxed vertical spring that has a spring constant of \(k=\) \(2.5 \mathrm{~N} / \mathrm{cm}\) (Fig. \(7-46)\). The block becomes attached to the spring and compresses the spring \(12 \mathrm{~cm}\) before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Particle Model of Matter

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Scientific Method Physics

Read Explanation

Energy Physics

Read Explanation

Engineering Physics

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.