# Work energy theorem

## Homework Statement

A moving electron has kinetic energy K1 . 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 K1

B) Does your answer depend o the final direction of the electrons motion? No

## The Attempt at a Solution

Hey guys so I learned how to solve this problem, but I dont really understand why they did what they did.

So in the problem they want to make everything in terms of K1

W= K2-k1
W=1/2mvf-1/2mvi

Here is where I get confused. The problem states that the electron is moving one quarter as fast in the opposite direction so wouldnt it be moving 1/4vf, and not 1/4vo?

For part B it was just yes or no, but I want to make sure I understand why it is no.

The reason why the answer does not depend on the final direction of the electrons motion is due to it being in terms of K1 and not in terms of K2.

Thank you

Related Introductory Physics Homework Help News on Phys.org
Work done by net external force is change in kinetic energy (Why?)

Change in kinetic energy is Kf-Ki.
Since velocity becomes 1/4th the kinetic energy become 1/16th (Why?)
Whats W now?

Answering your second question based on mathematical equation stated above will be easy. (Your current answer of No is right but reasoning is totally wrong.Dont worry its easy)

But can you explain why based on qualitative reasoning that why work done is independent of direction of motion?

When we say travelling with velocity 1/4th we mean the final velocity = initial velocity/4.

Edit
The mistake you are making is that vf=vi/4.

when your are saying it should be moving with vf/4 you are infact saying that vf=vf/4 which is wrong.

Makes perfect sense why it would be initial velocity over 4.

I don't really understand why work done would be independent of direction of motion without throwing in some equation.

mathematically,

net work done by net external force is change in kinetic energy.

Now kinetic energy is independent of direction of motion as it is a function of V squared (so it depends on speed squared ). As kinetic energy is independent , change in kinetic energy is also independent and so is work done by net external force

This is valid for all cases (at classical mechanics level) i.e work done by net external force is independent of direction of motion and depends only on change of kinetic energy(which in turn depends on magnitude of velocity i.e speed and not direction).

To understand qualitatively,

let us suppose a particle was travelling with 16 m/s towards the right .

As force is applied the new speed now become one fourth of the initial i.e 4 m/s.
its velocity however can be towards left or right.
Now we have to prove that in both cases work done by external force is same.

clearly for speed to go from 16 m/s to 4 m/s the force should act opposite to the motion i.e towards left

This force slowly decelerated the velocity to 4m/s (however the velocity is still towards the right).

Now the force could stop applying here for case 1 i.e body moving right with velocity 4 m/s.

or it could continue acting towards left making the velocity from +4m/s to 0 m/s to -4 m/s.

it turns out that the work done by force in changing the velocity from 4m/s to -4 m/s is 0(which we can see mathematically as well as change in kinetic energy is 0.).

this is because in going from +4 m/s to 0 m/sthe force acts towards left whereas the partcile still displaces towards the right(remember velocity is towards right and velocity is parallel to displacement vector), so work done is negative.
(work=force.displacement.cos(180)=-force*displacement)

however, after the body reaches 0 m/s, work done in changing velocity from 0m/s to -4m/s is positive as force acts towards left(it was always acting towards left).
Here the displacement is also towards left (as velocity is negative) so work comes out to be positive.(theta is 0 so cos0=1)

the net effect is that work done in the two cases cancels out.
making the net work done independent of direction of motion.

Thank you, I appreciate you taking the time out to thoroughly explain this concept.

No problem.Spreading the joy of knowledge gives me pleasure :-).
Have a good day.