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Chapter 7: Problem 52

How much work is required to stop an electron \(\left(m=9.11 \times 10^{-31} \mathrm{~kg}\right)\) which is moving with a speed of \(1.40 \times 10^{6} \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
The work required is \(-8.93 \times 10^{-19} \mathrm{~J}\).

Step by step solution

01

Understand Work-Energy Principle

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. Here, the work required to stop the electron will be equal to the negative of its initial kinetic energy since it will bring the electron to zero velocity.
02

Calculate Initial Kinetic Energy

Use the formula for kinetic energy, which is \[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass and \( v \) is the velocity. Substituting the given values for the electron:\[ KE = \frac{1}{2} \times 9.11 \times 10^{-31} \mathrm{~kg} \times (1.40 \times 10^6 \mathrm{~m/s})^2 \]
03

Perform the Calculations

Calculate the value of \( v^2 \) first:\[ (1.40 \times 10^6)^2 = 1.96 \times 10^{12} \]Now substitute back into the kinetic energy formula:\[ KE = \frac{1}{2} \times 9.11 \times 10^{-31} \times 1.96 \times 10^{12} \]\[ KE = 8.9318 \times 10^{-19} \mathrm{~J} \]
04

Interpret the Result

Since the work required to stop the electron is equal to the negative of its initial kinetic energy, the work done is \[ -8.9318 \times 10^{-19} \mathrm{~J} \]. The negative sign indicates that the work done is in the direction opposite to the electron's motion to bring it to rest.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It's a fundamental concept in physics that helps us understand how moving objects do work. The amount of kinetic energy an object has depends on two things: its mass and its velocity.
  • The formula used to calculate kinetic energy is: \[ KE = \frac{1}{2} m v^2 \] where \( m \) is the mass and \( v \) is the velocity.
  • Kinetic energy is always a positive number since it involves the square of the velocity.
  • An object that isn't moving has zero kinetic energy.
In the exercise, we calculate the kinetic energy of an electron moving at a high speed. We plug in its mass and speed into the formula to find out how much energy is required to stop it. This understanding of kinetic energy is crucial for a wide range of scientific and engineering applications.
Electron Dynamics
Electron dynamics relates to how electrons move and behave under certain conditions. Electrons are incredibly small particles, having a tiny mass but can move at very high velocities, sometimes close to the speed of light.
  • The motion of electrons is influenced by various forces they encounter.
  • In the given exercise, we're interested in stopping an electron, which involves understanding its current kinetic state and how much energy transfer is necessary to bring it to rest.
  • Electron dynamics often involve interactions with electric and magnetic fields, but in this simplified scenario, we're focused on the kinetic aspect.
This segment of physics is crucial for fields like electronics, where the control and movement of electrons form the basis for all circuit functions. Understanding electron dynamics aids in designing better and more efficient electronic devices.
Energy Calculations
Energy calculations involve using various equations and principles to determine the amount of energy associated with a process or state. When performing energy calculations, it's important to have accurate values for all variables involved.
  • The exercise showcases energy calculations by determining the work needed to bring an electron to rest. This involves the initial calculation of its kinetic energy.
  • These calculations need a step-by-step approach to ensure accuracy, as seen in the solution where we compute each part carefully before reaching the final energy value.
  • Energy calculations aren't limited to kinetic energy. They can include potential energy, thermal energy, and more, depending on the situation.
Precise energy calculations are vital across many scientific disciplines, helping us understand how systems operate and enabling innovations in technology and sustainability.

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Most popular questions from this chapter

An elevator cable breaks when a \(925-\mathrm{kg}\) elevator is \(22.5 \mathrm{~m}\) above the top of a huge spring \((k=\) \(\left.8.00 \times 10^{4} \mathrm{~N} / \mathrm{m}\right)\) at the bottom of the shaft. Calculate (a) the work done by gravity on the elevator before it hits the spring; (b) the speed of the elevator just before striking the spring; \((c)\) the amount the spring compresses (note that here work is done by both the spring and gravity).

A \(6.10 - \mathrm { kg }\) block is pushed 9.25\(\mathrm { m }\) up a smooth \(37.0 ^ { \circ }\) inclined plane by a horizontal force of 75.0\(\mathrm { N }\) . If the initial speed of the block is 3.25\(\mathrm { m } / \mathrm { s }\) up the plane, calculate \(( a )\) the initial kinetic energy of the block; \(( b )\) the work done by the 75.0 -N force; \(( c )\) the work done by gravity; \(( d )\) the work done by the normal force; \(( e )\) the final kinetic energy of the block.

What should be the spring constant \(k\) of a spring designed to bring a \(1300-\mathrm{kg}\) car to rest from a speed of \(90 \mathrm{~km} / \mathrm{h}\) so that the occupants undergo a maximum acceleration of \(5.0 \mathrm{~g}\) ?

Assume a cyclist of weight \(m g\) can exert a force on the pedals equal to \(0.90 \mathrm{mg}\) on the average. If the pedals rotate in a circle of radius \(18 \mathrm{~cm}\), the wheels have a radius of \(34 \mathrm{~cm}\), and the front and back sprockets on which the chain runs have 42 and 19 teeth respectively (Fig. \(7-31\) ), determine the maximum stecpness of hill the cyclist can climb at constant speed. Assume the mass of the bike is \(12 \mathrm{~kg}\) and that of the rider is \(65 \mathrm{~kg} .\) Ignore friction. Assume the cyclist's average force is always: \((a)\) downward; \((b)\) tangential to pedal motion.

Consider a force \(F_{1}=A / \sqrt{x}\) which acts on an object during its journey along the \(x\) axis from \(x=0.0\) to \(x=1.0 \mathrm{~m},\) where \(A=2.0 \mathrm{~N} \cdot \mathrm{m}^{1 / 2},\) Show that during this journey, even though \(F_{1}\) is infinite at \(x=0.0\), the work done on the object by this force is finite.
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