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Kinetic Energy of bullet fired from gun

  1. Aug 17, 2014 #1
    Hi everyone. I'm quite troubled with this physics question.

    A .035 kg bullet is fired from a .23 m barrel. The bullet experiences a force of 4500 N while in the gun barrel. What is the kinetic energy of the bullet as it leaves the gun barrel?

    I don't understand how to figure out the velocity to calculate the KE. Is it possible to calculate the KE from GPE or Work?

    Our teacher has not taught us this and I don't understand why he is expecting us to answer this question....
    (btw. I'm in year 10)

    Thanks.
     
  2. jcsd
  3. Aug 17, 2014 #2

    Doc Al

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    Yes. Look up the work-energy theorem. (See: Work-Energy Principle)

    Calculate the work done by that force. (Assume that it's constant.)
     
  4. Aug 17, 2014 #3
    So, If I assume It is constant. The answer = 4500 N x .23 m = 1035 J?
     
    Last edited: Aug 17, 2014
  5. Aug 17, 2014 #4

    Doc Al

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    Yep. You got it.
     
  6. Aug 17, 2014 #5

    Thanks. I don't understand why our teacher gave us this question when we hadn't learnt the work energy theorem... Anyways thanks.
     
  7. Aug 17, 2014 #6

    CWatters

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    You could also solve it by applying Newtons laws and equations of motion.
     
  8. Aug 17, 2014 #7

    We have not learned equations of motion. I have no clue about them.
     
  9. Aug 17, 2014 #8

    BvU

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    Pity you didn't use the template. Under 2) relevant equations you would have filled in something, right? I mean by the time you are in year 10 there must have been something comining by that can be related to this exercise ?
    What would you have filled in if you absolutely would have needed to fill in something ?
     
  10. Aug 18, 2014 #9
    Classics :

    force (N) = mass (kg) * ( constant ) acceleration ( m/s/s )

    You have mass ( 0.035 kg ) and force ( 4,500 N ), so, transpose equation for acceleration :
    acceleration = force / mass
    acceleration = 128,571.43 m/s/s

    With the acceleration (a) and distance (s), you can use the classic equation :
    v ² = u ² + ( 2 * a * s )
    u = initial velocity which = 0, so drop it, then transpose for v ( final velocity )
    So :
    v = square root ( 2 * a * s )
    v = 243.193 m/s

    Now find the KE from :
    KE = ½ * mass * velocity ²
    KE = 1,035 Joules
     
  11. Aug 18, 2014 #10

    BvU

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    I'm still wondering if Bob had an empty toolbag to begin with, or if there was something he could use to work out this exercise...
     
  12. Aug 18, 2014 #11
    Umm yeah Sorry. I didn't realise they were 3 columns that we could fill in. Because with my second question I asked on another thread I realised. :/
     
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