Motion along a line due to a constant net force

  • Thread starter .NoStyle
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  • #1
.NoStyle
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Homework Statement



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?



Homework Equations



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



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:

Answers and Replies

  • #2
Ed Aboud
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Hey
I'm not really clear on what you are trying to do but try using [tex] F = ma [/tex].
 
  • #3
.NoStyle
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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:
  • #4
Ed Aboud
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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
.NoStyle
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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
Ed Aboud
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Ok so read over the question again, it says that it stops accelerating after 2cm, right?
 
  • #7
.NoStyle
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it stops accelerating after 2.0cm and then continues at a constant velocity for 45cm.

Thanks Ed
 
  • #8
Ed Aboud
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Recheck your calculation of the acceleration.
 
  • #9
Ed Aboud
199
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No problem.
:smile::smile:
 
  • #10
.NoStyle
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hmm, I still get 1.00128E-21 m/s^2 for acceleration...
 

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