Quick Question - Explain Work Energy Concept

In summary, the conversation discusses using the work-energy theorem to find the force required to accelerate an electron from rest to a specific speed in a given distance. The work-energy theorem states that work equals the change in kinetic energy, and in this case, the electron starts from rest and has no initial kinetic energy. By setting the work equal to force times distance, the force can be calculated using the given mass and velocity of the electron.
  • #1
physicsmarkb
4
0
[Solved] Quick Question - Explain Work Energy Concept

Homework Statement



Use the work - kinetic energy theorem to find the force required to accelerate an electron (m = 9.11 x 10^-31 kg) from rest to a speed of 1.50 x 10^7 m/s in a distance of .0125 m.


Homework Equations


W = Fx cos(theta), K = 1/2 mv^2


The Attempt at a Solution



I know that I would set Fx cos(theta) = 1/2 mv^2 and then solve for F but I don't understand why I would do this.
 
Last edited:
Physics news on Phys.org
  • #2
Does anyone know why I have have to solve it like this?
 
  • #3
work = the change in kinetic energy, in other words the final kinetic energy - the initial kinetic energy. The electron starts from rest and has no initial kinetic energy. That pretty much IS the work-energy theorem

Make sure you understand what work is, it's how much energy you're going to impart to the electron, and you know what final kinetic energy you would like the electron to have
 
  • #4
So with K final, all you have is mass and velocity. With W = Fx, you have force and distance. Why do you divide mass and velocity by distance?
Edit: I just realized that you divide because you're trying to solve for F.

What I still don't understand is that why you have all the information for K = 1/2 mv^2 which is W = K but that doesn't work. I don't get why you have to set it equal to W = Fx.
 
Last edited:
  • #5
What I still don't understand is that why you have all the information for K = 1/2 mv^2 which is W = K but that doesn't work. I don't get why you have to set it equal to W = Fx.

Well with W=K and knowing K, you would've found work. Were you looking for work? No. Can you use the definition of work to find force? Yes. And that's what you did
 
  • #6
blochwave said:
Well with W=K and knowing K, you would've found work. Were you looking for work? No. Can you use the definition of work to find force? Yes. And that's what you did

Wow I'm so dumb. I can't believe I wasn't paying attention to W and F.
 

1. What is the work-energy concept?

The work-energy concept is a fundamental principle in physics that states that the work done on an object is equal to the change in its kinetic energy. In other words, work is the transfer of energy from one form to another.

2. How is work calculated in the work-energy concept?

Work is calculated by multiplying the force applied to an object by the distance over which the force is applied. The unit of work is joules (J).

3. What is the relationship between work and energy in the work-energy concept?

The work-energy concept states that work done on an object will result in a change in its energy. This change in energy can be in the form of kinetic energy, potential energy, or a combination of both.

4. How does the work-energy concept apply to real-life situations?

The work-energy concept is applicable in various real-life situations, such as pushing a cart, lifting an object, or riding a bike. In each case, work is done on the object, resulting in a change in its energy.

5. Can the work-energy concept be applied to non-conservative forces?

Yes, the work-energy concept can be applied to non-conservative forces, but the work done by these forces will not result in a change in the object's total mechanical energy. This is because non-conservative forces can dissipate energy, such as friction or air resistance.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
468
  • Introductory Physics Homework Help
Replies
5
Views
764
  • Introductory Physics Homework Help
2
Replies
56
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
911
  • Introductory Physics Homework Help
Replies
7
Views
214
  • Introductory Physics Homework Help
Replies
4
Views
819
  • Introductory Physics Homework Help
Replies
5
Views
175
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
2
Replies
55
Views
2K
Replies
11
Views
1K
Back
Top