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A moving electron has kinetic energy \(K_{1}\) . After a net amount of work \(W\) has been done on it, the electron is moving one-quarter as fast in the opposite direction. (a) Find \(W\) in terms of \(K_{1}\) . (b) Does your answer depend on the final direction of the electron's motion?

Short Answer

Expert verified
(a) \( W = -\frac{15}{16} K_1 \); (b) No, direction does not affect \( W \).

Step by step solution

01

Understand the Initial Situation

Initially, the electron has a kinetic energy given by \( K_1 = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron and \( v \) is its velocity. This describes the kinetic energy before any work is done on it.
02

Define the Final Situation

After work \( W \) is done, the electron's velocity becomes \( -\frac{v}{4} \). This indicates that its final speed is one-quarter of the initial speed, but in the opposite direction. The new kinetic energy \( K_2 \) is \( \frac{1}{2}m\left(-\frac{v}{4}\right)^2 = \frac{1}{2}m\frac{v^2}{16} = \frac{1}{32} mv^2 \).
03

Relate Work Done to Change in Kinetic Energy

By the work-energy principle, the net work done on the electron is equal to the change in its kinetic energy. Therefore, \( W = K_2 - K_1 = \frac{1}{32}mv^2 - \frac{1}{2}mv^2 \).
04

Express Work in Terms of Initial Kinetic Energy

Given that \( K_1 = \frac{1}{2}mv^2 \), we can express \( W \) as follows: \( W = -\frac{15}{32}mv^2 \). Since \( K_1 = \frac{1}{2}mv^2 \), we replace \( \frac{1}{2}mv^2 \) with \( K_1 \) in the equation, obtaining \( W = -\frac{15}{16} K_1 \).
05

Consider Direction Dependence

Since kinetic energy is a scalar quantity, it does not depend on the direction of motion. The work done, as calculated from changes in kinetic energy, does not depend on the electron's change in direction.

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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 a crucial concept when discussing the motion of any object, including tiny particles like electrons.
It is defined as the energy that an object possesses because of its motion. When an electron moves, it exhibits kinetic energy, which can be calculated using the formula:
  • \( K = \frac{1}{2}mv^2 \)
Here, \( m \) is the mass of the electron, and \( v \) is its velocity.
This tells us that the kinetic energy is directly proportional to the mass and the square of the velocity of the object. In our exercise, the initial kinetic energy of the electron is \( K_1 \).
After some work is done on it, its velocity changes, leading to a new kinetic energy \( K_2 \). The work-energy principle helps us understand how this change occurs by stating that the work done on the electron alters its kinetic energy.
Electron Dynamics
Electron dynamics involves understanding how electrons move under the influence of forces such as electric fields or other forces doing work.
When forces act on electrons, their motion changes, which can be represented by a modification in their velocity, and consequently, their kinetic energy changes as well.When we apply the work-energy principle, it becomes clear how work impacts an electron's velocity and kinetic energy.
  • Initially, the electron has a certain speed with its corresponding kinetic energy, \( K_1 \).
  • Once a specified work \( W \) is done, the electron's velocity diminishes by a factor of one-fourth while also reversing direction, resulting in a new kinetic energy value, \( K_2 \).
This exercise exemplifies how the work done on an electron directly modifies its motion dynamics, demonstrating the powerful connection between external work and changes in an electron's velocity and kinetic energy.
Direction Independence
When analyzing motion, you might wonder if direction plays a role in determining work and energy.
Interestingly, kinetic energy is a scalar quantity, which means it doesn't depend on direction.
  • Whether an electron moves left, right, or changes its direction entirely, its kinetic energy value remains determined solely by its speed and mass.
  • This concept helps simplify our calculations since the work-energy principle only considers the magnitude of speed changes, not directions.
In the given problem, the change in direction of the electron, moving in the opposite way with reduced speed, shows that while the direction altered, the net work done on the electron remains unaffected.
This provides a clear example of direction independence in energy calculations, which is a helpful guideline in physics when considering other force and motion scenarios.

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Most popular questions from this chapter

A Near-Earth Asteroid. On April \(13,2029\) (Friday the 13 th!), the asteroid 99942 Apophis will pass within \(18,600 \mathrm{mi}\) of the earth-about 1\(/ 13\) the distance to the moon! It has a density of 2600 \(\mathrm{kg} / \mathrm{m}^{3}\) , can be modeled as a sphere 320 \(\mathrm{m}\) in diameter, and will be traveling at 12.6 \(\mathrm{km} / \mathrm{s}\) . (a) If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver? (b) The largest nuclear bomb ever tested by the United States was the "CastlefBravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases \(4.184 \times 10^{15} \mathrm{J}\) of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?

Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount \(x\) , a force along the \(x\) -axis with \(x\)-component \(F_{x}=k x-b x^{2}+c x^{3}\) must be applied to the free end. Here \(k=100 \mathrm{N} / \mathrm{m}, b=700 \mathrm{N} / \mathrm{m}^{2},\) and \(c=12,000 \mathrm{N} / \mathrm{m}^{3}\). Note that \(x>0\) when the spring is stretched and \(x<0\) when it is compressed. (a) How much work must be done to stretch this spring by 0.050 \(\mathrm{m}\) from its unstretched length? (b) How much work must be done to compress this spring by 0.050 \(\mathrm{m}\) from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of \(F_{x}\) on \(x\) . (Many real springs behave qualitatively in the same way.)

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \(\vec{F}=(30 \mathrm{N}) \hat{t}-(40 \mathrm{N}) \hat{\jmath}\) to the cart as it undergoes a displacement \(\vec{s}=(-9.0 \mathrm{m}) \hat{\imath}-(3.0 \mathrm{m}) \hat{\mathrm{j}}\) How much work does the force you apply do on the grocery cart?

You are asked to design spring bumpers for the walls of a parking garage. A freely rolling \(1200-\mathrm{kg}\) car moving at 0.65 \(\mathrm{m} / \mathrm{s}\) is to compress the spring no more than 0.070 \(\mathrm{m}\) before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass.

A tow truck pulls a car 5.00 \(\mathrm{km}\) along a horizontal roadway using a cable having a tension of 850 \(\mathrm{N}\) . (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part \((a) ?(c)\) How much work does gravity do on the car in part (a)?

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