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Chapter 7: Problem 44

Compare the kinetic energy of a 20,000 -kg truck moving at \(110 \mathrm{km} / \mathrm{h}\) with that of an \(80.0-\mathrm{kg}\) astronaut in orbit moving at \(27,500 \mathrm{km} / \mathrm{h}\).

Short Answer

Expert verified
The kinetic energy of the \(20,000\)-kg truck moving at \(30.56 \frac{\text{m}}{\text{s}}\) is \(9,405,120 \mathrm{J}\), while the kinetic energy of the \(80.0\)-kg astronaut moving at \(7,638.89 \frac{\text{m}}{\text{s}}\) is \(23,325,720,320 \mathrm{J}\). The astronaut has a significantly higher kinetic energy than the truck due to their much higher velocity, despite having a smaller mass.

Step by step solution

01

Convert velocities to m/s

We first need to convert the velocities from km/h to m/s. We can do this using the following conversion: \[1 \mathrm{km}/\mathrm{h} = \frac{1000 \mathrm{m}}{3600 \mathrm{s}}\] For the truck: \[v_{truck} = 110 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 30.56 \frac{\mathrm{m}}{\mathrm{s}}\] For the astronaut: \[v_{astronaut} = 27,500 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 7,638.89 \frac{\mathrm{m}}{\mathrm{s}}\]
02

Calculate the kinetic energies

Now we'll use the formula for kinetic energy and the calculated velocities to determine the kinetic energies of both the truck and the astronaut. For the truck: \[KE_{truck} = \frac{1}{2}m_{truck}v_{truck}^2 = \frac{1}{2}(20,000 \mathrm{kg})(30.56 \frac{\mathrm{m}}{\mathrm{s}})^2 = 9,405,120 \mathrm{J}\] For the astronaut: \[KE_{astronaut} = \frac{1}{2}m_{astronaut}v_{astronaut}^2 = \frac{1}{2}(80 \mathrm{kg})(7,638.89 \frac{\mathrm{m}}{\mathrm{s}})^2 = 23,325,720,320 \mathrm{J}\]
03

Compare the kinetic energies

Let's now compare the calculated kinetic energies: \[KE_{truck} = 9,405,120 \mathrm{J} \quad and \quad KE_{astronaut} = 23,325,720,320 \mathrm{J}\] As we can see, the astronaut has a significantly higher kinetic energy than the truck, despite having a much smaller mass. This is mainly due to the significantly higher velocity of the astronaut compared to the truck.

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Key Concepts

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

Kinetic Energy Formula
Kinetic energy is a form of energy that an object possesses due to its motion. The kinetic energy of an object can be calculated using the kinetic energy formula:

\[\begin{equation} KE = \frac{1}{2}mv^2 \end{equation}\]

Where:
  • KE represents the kinetic energy in joules (J)
  • m is the mass of the object in kilograms (kg)
  • v is the velocity of the object in meters per second (m/s)

It is crucial to understand that both mass and velocity contribute to an object's kinetic energy, but velocity has a greater influence since it is squared in the formula. This means even a small increase in velocity will result in a much larger increase in kinetic energy. For example, in our textbook problem, the astronaut's kinetic energy is vastly greater than the truck's, primarily because of the astronaut's much higher velocity.
Unit Conversion
Unit conversion is a critical step in many physics problems as it ensures that all quantities are expressed in compatible units before calculations are performed.

In our exercise, we convert velocity from kilometers per hour (km/h) to meters per second (m/s) using the conversion factor:
\[\begin{equation} 1 \mathrm{km}/\mathrm{h} = \frac{1000 \mathrm{m}}{3600 \mathrm{s}} \end{equation}\]

Understanding and applying the correct unit conversion can greatly affect the outcome of a calculation. By expressing velocities in m/s, we obtain values that contribute directly to the kinetic energy formula, which requires velocity in the SI (International System of Units) standard unit of m/s. The ability to accurately convert units is not only essential in solving physics problems but is also a valuable skill in everyday situations where various measurements are encountered.
Velocity in Physics
In physics, velocity is a vector quantity that describes both the speed of an object and its direction of motion. Velocity is often expressed in terms of units such as meters per second (m/s) or kilometers per hour (km/h). It is an essential concept because it allows us to understand how the position of an object changes over time.

Velocity in physics is not just about how fast something is moving; it's also about where it's heading. The square of velocity plays a significant role in kinetic energy as seen in the kinetic energy formula. This was further illustrated in our textbook problem where the astronaut's tremendous velocity led to much higher kinetic energy compared to the truck鈥攈ighlighting that even objects with much lesser mass can have exceptional kinetic energy if their velocity is sufficiently high.

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Most popular questions from this chapter

A sled plus passenger with total mass \(50 \mathrm{kg}\) is pulled \(20 \mathrm{m}\) across the snow \(\left(\mu_{k}=0.20\right)\) at constant velocity by a force directed \(25^{\circ}\) above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.

An electron in a television tube is accelerated uniformly from rest to a speed of \(8.4 \times 10^{7} \mathrm{m} / \mathrm{s}\) over a distance of \(2.5 \mathrm{cm} .\) What is the power delivered to the electron at the instant that its displacement is \(1.0 \mathrm{cm} ?\)

A 5.0-kg box has an acceleration of \(2.0 \mathrm{m} / \mathrm{s}^{2}\) when it is pulled by a horizontal force across a surface with \(\mu_{K}=0.50 .\) Find the work done over a distance of \(10 \mathrm{cm}\) by (a) the horizontal force, (b) the frictional force, and (c) the net force. (d) What is the change in kinetic energy of the box?

Engineers desire to model the magnitude of the elastic force of a bungee cord using the equation $$ F(x)=a\left[\frac{x+9 m}{9 m}-\left(\frac{9 m}{x+9 m}\right)^{2}\right] $$ where \(x\) is the stretch of the cord along its length and \(a\) is a constant. If it takes \(22.0 \mathrm{kJ}\) of work to stretch the cord by 16.7 m, determine the value of the constant \(a\)

Using energy considerations, calculate the average force a 60.0 -kg sprinter exerts backward on the track to accelerate from 2.00 to \(8.00 \mathrm{m} / \mathrm{s}\) in a distance of \(25.0 \mathrm{m}\), if he encounters a headwind that exerts an average force of 30.0 N against him.
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