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Chapter 4: Problem 57

An electron \(\left(\mathrm{mass}=9.11 \times 10^{-31} \mathrm{kg}\right)\) leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is 1.80 \(\mathrm{cm}\) away. It reaches the grid with a speed of \(3.00 \times 10^{6} \mathrm{m} / \mathrm{s} .\) If the accelerating force is constant, compute (a) the acceleration of the electron, (b) the time it takes the electron to reach the grid, and (c) the net force that is accelerating the electron, in newtons. (You can ignore the gravitational force on the electron.)

Short Answer

Expert verified
(a) \(a \approx 2.5 \times 10^{14} \, \text{m/s}^2 \), (b) \(t \approx 1.2 \times 10^{-8} \, \text{s} \), (c) \(F \approx 2.28 \times 10^{-16} \, \text{N} \).

Step by step solution

01

Identify Given Variables

Start by listing the information given in the problem: - Initial velocity \( u = 0 \, \text{m/s} \)- Final velocity \( v = 3.00 \times 10^6 \, \text{m/s} \)- Distance \( s = 1.80 \, \text{cm} = 0.018 \, \text{m} \)- Mass of electron \( m = 9.11 \times 10^{-31} \, \text{kg} \)
02

Use Kinematic Equation to Find Acceleration

Use the kinematic equation: \[ v^2 = u^2 + 2as \]Substitute the known values:\[ (3.00 \times 10^6 \, \text{m/s})^2 = 0^2 + 2 \cdot a \cdot 0.018 \]\[ 9.00 \times 10^{12} = 2a \cdot 0.018 \]Solve for \( a \):\[ a = \frac{9.00 \times 10^{12}}{0.036} \approx 2.5 \times 10^{14} \, \text{m/s}^2 \]
03

Calculate Time Using Another Kinematic Equation

Use another kinematic equation:\[ v = u + at \]Substitute the known values:\[ 3.00 \times 10^6 = 0 + (2.5 \times 10^{14})t \]Solve for \( t \):\[ t = \frac{3.00 \times 10^6}{2.5 \times 10^{14}} \approx 1.2 \times 10^{-8} \, \text{s} \]
04

Calculate Net Force Using Newton's Second Law

Apply Newton's second law:\[ F = ma \]Substitute the known values:\[ F = 9.11 \times 10^{-31} \times 2.5 \times 10^{14} \]\[ F \approx 2.28 \times 10^{-16} \, \text{N} \]

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Key Concepts

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

Kinematic Equations
Kinematic equations are essential tools for solving problems involving motion, such as the movement of an electron in a TV picture tube.
These equations relate variables such as displacement, velocity, acceleration, and time.
When solving these problems, you can use kinematic equations to find unknowns if you know some of the other values.For the electron exercise, two key kinematic equations were employed:
  • The first equation, \( v^2 = u^2 + 2as \), helps find the acceleration of the electron moving in the tube, given its initial and final velocities, and the distance traveled.
  • The second equation, \( v = u + at \), determines the time taken by using the calculated acceleration.
Understanding the relationships between these variables allows you to plug in the given values and solve the problem step by step.
Newton's Second Law
Newton's second law is a fundamental principle in physics.
It explains the relationship between an object's mass, the force applied to it, and the resulting acceleration.
According to this law, the force acting on an object is equal to its mass multiplied by its acceleration. This is expressed as:\[ F = ma \]In the problem of the electron, once the acceleration was calculated using the kinematic equations, applying Newton's second law was straightforward.
By substituting the mass of the electron and the computed acceleration into the formula, you calculate the net force acting on the electron.
This showcases how intertwining different physics principles can help solve complex problems.
Initial Velocity
Understanding initial velocity is crucial when analyzing motion.
It represents the speed and direction of an object at the beginning of its motion. In many problems, especially those involving acceleration, the initial velocity will be key in setting up calculations.In the scenario with the electron traveling in a TV picture tube, the initial velocity \( u \) is given as zero.
This simplifies the problem significantly since many kinematic equations require consideration of both initial and final velocities.
With \( u = 0 \), calculations for acceleration and time become more straightforward, as terms involving the initial velocity vanish.
This concept highlights how choosing the right starting point simplifies complex calculations.
Final Velocity
Final velocity is another critical aspect of kinematics, representing the speed and direction of an object at the end of its motion.
It is a vital piece of information in predicting or analyzing an object's behavior after being subjected to forces.In the electron problem, the final velocity is provided as \( 3.00 \times 10^6 \mathrm{m/s} \).
With this value, you can reverse engineer other motion variables such as acceleration and time, using the kinematic equations.
The known final velocity, alongside calculated acceleration, helps solve for time in which the electron reaches this speed.
This demonstrates how mastering these fundamental concepts can make tackling physics problems more manageable and less intimidating.

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Most popular questions from this chapter

A standing vertical jump. Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 \(\mathrm{m}\) (4 ft). (This means that he moved upward by 1.2 \(\mathrm{m}\) after his feet left the floor.) Griffith weighed 890 \(\mathrm{N}(200 \mathrm{lb}) .\) (a) What was his speed as he left the floor? (b) If the time of the part of the jump before his feet left the floor was 0.300 s, what were the magnitude and direction of his acceleration (assuming it to be constant) while he was pushing against the floor? (c) Draw a free-body diagram of Griffith during the jump. (d) Use Newton's laws and the results of part (b) to calculate the force he applied to the ground during his jump.

An athlete whose mass is 90.0 \(\mathrm{kg}\) is performing weight lifting exercises. Starting from the rest position, he lifts, with constant acceleration, a barbell that weighs 490 \(\mathrm{N} .\) He lifts the barbell a distance of 0.60 \(\mathrm{m}\) in 1.6 \(\mathrm{s}\) . (a) Draw a clearly labeled free-body force diagram for the barbell and for the athlete. (b) Use the diagrams in part (a) and Newton's laws to find the total force that the ground exerts on the athlete's feet as he lifts the barbell.

Human biomechanics. The fastest served tennis ball, served by "Big Bill" Tilden in \(1931,\) was measured at 73.14 \(\mathrm{m} / \mathrm{s}\) . The mass of a tennis ball is \(57 \mathrm{g},\) and the ball is typically in contact with the tennis racquet for 30.0 \(\mathrm{ms}\) , with the ball starting from rest. Assuming constant acceleration, (a) what force did Big Bill's tennis racquet exert on the tennis ball if he hit it essentially horizontally? (b) Make free- body diagrams of the tennis ball during the serve and just after it has moved free of the racquet.

Human biomechanics. The fastest pitched baseball was measured at 46 \(\mathrm{m} / \mathrm{s}\) . Typically, a baseball has a mass of 145 \(\mathrm{g}\) . If the pitcher exerted his force (assumed to be horizontal and constant) over a distance of \(1.0 \mathrm{m},(\mathrm{a})\) what force did he produce on the ball during this record-setting pitch? (b) Make free-body diagrams of the ball during the pitch and just after it has left the pitcher's hand.

Superman throws a 2400 -N boulder at an adversary. What horizontal force must Superman apply to the boulder to give it a horizontal acceleration of 12.0 \(\mathrm{m} / \mathrm{s}^{2} ?\)
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