# Motion along a line due to a constant net force

1. Jun 14, 2008

### .NoStyle

1. The problem statement, all variables and given/known data

In a cathode ray tube, electrons are accelerated from rest by a constant electric force of magnitude 6.4E-17 N during the first 2.0cm of the tubes length; then they move at essentially constant velocity another 45cm before hitting the screen.

A) Find the speed of the electrons when they hit the screen.

B) How long does it take them to travel the length of the tube?

2. Relevant equations

Delta(v)subx = Vsub(f)x - vsub(i)(x) = asubx(delta(t))

3. The attempt at a solution

well, we can reduce the equation to:

Delta(v)subx = asubx(delta(t))

but then I only have a force in Newtons, and a distance of 2.0cm, then another 45cm at a constant velocity. So I don't even know how I can derive anything like Delta(t) or asubx.

Thank you

Last edited: Jun 14, 2008
2. Jun 14, 2008

### Ed Aboud

Hey
I'm not really clear on what you are trying to do but try using $$F = ma$$.

3. Jun 14, 2008

### .NoStyle

hi ed, so I know the force which is:

6.4E-17 N = m*a

I can google the mass of an electron, but should I really have to to solve this problem? Since the mass of an electron isn't given in the book?

(mass of electron = 9.10938188 × 10-31 kg in google search).

so if I do your way:

6.4XE-17 N / 9.10938188E-31 = acceleration ???

Since newtons is essentially kg*m/s^2 right? Thanks

edit:

So I get 1.00128E-21m/S^2 for acceleration

Last edited: Jun 14, 2008
4. Jun 14, 2008

### Ed Aboud

Yep, thats right.
Hmm it's strange that the mass wasn't given to you.
But to be honest I can't see anything wrong with this way.
If I was presented with this question I would do it that way.

5. Jun 14, 2008

### .NoStyle

Ed, that makes sense.

So I find the acceleration is 1.00128E-21m/s^2

I convert that to centimeters and get:

1.00128E-19 cm/s^2 = acceleration

so what do I do to solve question A which asks "Find the speed of the electrons when they hit the screen." ???

I have no clue what to do, any hints would be helpful. Thank you

6. Jun 14, 2008

### Ed Aboud

Ok so read over the question again, it says that it stops accelerating after 2cm, right?

7. Jun 14, 2008

### .NoStyle

it stops accelerating after 2.0cm and then continues at a constant velocity for 45cm.

Thanks Ed

8. Jun 14, 2008

### Ed Aboud

Recheck your calculation of the acceleration.

9. Jun 14, 2008

### Ed Aboud

No problem.

10. Jun 14, 2008

### .NoStyle

hmm, I still get 1.00128E-21 m/s^2 for acceleration...