/*! 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 3 In a TV picture tube electrons a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 3

In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of \(5 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in travelling one centimeter. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is \(9 \cdot 1 \times 10^{-31} \mathrm{~kg}\).

Short Answer

Expert verified
The constant force exerted on the electron is \(1.1375 \times 10^{-15} \text{ N}\).

Step by step solution

01

Identify Given Information

We are given the final velocity of the electron, \( v_f = 5 \times 10^{6} \, \mathrm{m/s} \), the initial velocity \( v_i \approx 0 \, \mathrm{m/s} \), the distance traveled \( s = 0.01 \, \mathrm{m} \), and the mass of the electron \( m = 9.1 \times 10^{-31} \, \mathrm{kg} \).
02

Use the Kinematic Equation

Use the kinematic equation \( v_f^2 = v_i^2 + 2as \) to solve for the acceleration \( a \). Since \( v_i \approx 0 \), the equation simplifies to \( v_f^2 = 2as \).
03

Solve for Acceleration

Rearrange the equation to solve for \( a \), giving \( a = \frac{v_f^2}{2s} \). Substitute the given values: \[ a = \frac{(5 \times 10^{6} \, \mathrm{m/s})^2}{2 \times 0.01 \, \mathrm{m}} = \frac{25 \times 10^{12} \, \mathrm{m^2/s^2}}{0.02 \, \mathrm{m}} = 1.25 \times 10^{15} \, \mathrm{m/s^2} \].
04

Use Newton's Second Law

According to Newton's second law, \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. Substitute the known values into the equation to find the force.
05

Calculate the Force

With \( m = 9.1 \times 10^{-31} \, \mathrm{kg} \) and \( a = 1.25 \times 10^{15} \, \mathrm{m/s^2} \), the force is \( F = (9.1 \times 10^{-31} \, \mathrm{kg})(1.25 \times 10^{15} \, \mathrm{m/s^2}) = 1.1375 \times 10^{-15} \, \mathrm{N} \).

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.

Kinematic Equations
Kinematic equations are fundamental in physics for describing the motion of objects. They link variables like velocity, acceleration, time, and displacement without needing to consider the forces involved. In the context of this exercise, we use one specific equation:
\[ v_f^2 = v_i^2 + 2as \]
This equation is perfect for scenarios where you know the initial velocity \( v_i \), final velocity \( v_f \), the distance traveled \( s \), and you need to find acceleration \( a \). It's especially useful when motion occurs in a straight line with constant acceleration.
  • Initial and Final Velocities: Here, the initial velocity \( v_i \) is approximately 0 as electrons start from rest. The final velocity \( v_f \) is given as \( 5 \times 10^{6} \, \text{m/s} \).
  • Distance: Electrons travel a distance of 0.01 meters.
  • Acceleration: Solving for \( a \), the equation simplifies to \( a = \frac{v_f^2}{2s} \).
Newton's Second Law
Newton's Second Law of Motion is a cornerstone of physics, explaining how the velocity of an object changes when it is subject to an external force. The law is expressed with the formula:
\[ F = ma \]
This demonstrates the relationship between force \( F \), mass \( m \), and acceleration \( a \). It's crucial for calculating how much force is needed to move an object at a specific acceleration.
In this exercise:
  • Mass: The mass of the electron is a tiny, constant value \( 9.1 \times 10^{-31} \, \text{kg} \).
  • Acceleration: Calculated previously as \( 1.25 \times 10^{15} \, \text{m/s}^2 \).
  • Force: Using the formula, \( F = (9.1 \times 10^{-31} \, \text{kg})(1.25 \times 10^{15} \, \text{m/s}^2) \), we find the force exerted on the electron.
Electron Acceleration
Electron acceleration is a vital concept for understanding electron dynamics. It's the rate of change of the velocity of electrons as they move through the TV picture tube, influenced by an electric field. Acceleration \( a \) is calculated using the equation derived from kinematics mentioned earlier.
  • Calculation: We've obtained \( a = 1.25 \times 10^{15} \, \text{m/s}^2 \) by rearranging the kinematic equation to solve for \( a \).
  • Significance: This high acceleration value illustrates how quickly an electron speeds up as it travels through a tiny distance in a controlled environment.
  • Application: The acceleration informs us about the necessary force using Newton's Second Law, crucial for understanding electron behavior in electronics.
Force Calculation
Force calculation is key to understanding how much effort is needed to move an electron through a TV picture tube, using previous concepts like Newton's Second Law. Once we have the acceleration, force can be calculated directly, providing insights into electron dynamics.
  • Given Mass: The mass of an electron is minuscule, but essential for calculating force.
  • Utilized Acceleration: The previously found acceleration \( 1.25 \times 10^{15} \, \text{m/s}^2 \) is used here.
  • Formula Application: Substitute \( m \) and \( a \) into \( F = ma \). The result is \( 1.1375 \times 10^{-15} \, \text{N} \), representing the force exerted on the electron.
  • Importance: Understanding this force gives insight into the interaction between electrical fields and charged particles.

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 monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end (figure 5 -E19). Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration. If initially both were at rest, their separation will not change as time passes.

A pendulum bob of mass \(50 \mathrm{~g}\) is suspended from the ceiling of an elevator. Find the tension in the string if the elevator (a) goes up with acceleration \(1 \cdot 2 \mathrm{~m} / \mathrm{s}^{2}\), (b) goes up with deceleration \(1 \cdot 2 \mathrm{~m} / \mathrm{s}^{2},(\mathrm{c})\) goes up with uniform velocity, (d) goes down with acceleration \(1 \cdot 2 \mathrm{~m} / \mathrm{s}^{2},(\mathrm{e})\) goes down with deceleration \(1 \cdot 2 \mathrm{~m} / \mathrm{s}{ }^{2}\) and (f) goes down with uniform velocity.

Two blocks \(A\) and \(B\) of mass \(m_{A}\) and \(m_{B}\) respectively are kept in contact on a frictionless table. The experimenter pushes the block \(A\) from behind so that the blocks accelerate. If the block \(A\) exerts a force \(F\) on the block \(B\), what is the force exerted by the experimenter on \(A\) ?

A block of \(2 \mathrm{~kg}\) is suspended from the ceiling through a massless spring of spring constant \(k=100 \mathrm{~N} / \mathrm{m} .\) What is the elongation of the spring ? If another \(1 \mathrm{~kg}\) is added to the block, what would be the further elongation?

The force of buoyancy exerted by the atmosphere on a balloon is \(B\) in the upward direction and remains constant. The force of air resistance on the balloon acts opposite to the direction of velocity and is proportional to it. The balloon carries a mass \(M\) and is found to fall down near the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity \(v ?\)
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Dynamics

Read Explanation

Energy Physics

Read Explanation

Linear Momentum

Read Explanation

Waves Physics

Read Explanation

Fields in 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.