Quick Question - Explain Work Energy Concept

AI Thread Summary
The discussion focuses on using the work-kinetic energy theorem to determine the force needed to accelerate an electron from rest to a specified speed over a given distance. Participants clarify that work is defined as the change in kinetic energy, which can be expressed through the equation W = Fx. The confusion arises from understanding why W must equal Fx when the kinetic energy formula K = 1/2 mv^2 provides the necessary information. Ultimately, the key takeaway is that while the kinetic energy can be calculated, the force is derived from the work-energy relationship. Understanding these concepts is crucial for solving the problem effectively.
physicsmarkb
Messages
4
Reaction score
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
Does anyone know why I have have to solve it like this?
 
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
 
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:
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
 
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top