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Homework Help: Laws of motion HELP

  1. Jun 8, 2005 #1
    Been struggling wih this problem all afternoon.

    On an aircraft carrier, a jet can be catapulted fom 0 to 155 mi/hr in 2.00 sec. If the average force exerted by the catapult is 5.10 * 10^6 Newtons, what is he mass of the jet?

    We converted the velocity into meters/second and got 69.276 meters/second

    initial velocity = 0
    initial X = 0
    final X = (initial X) + (velocity of X)(time)
    = 0 + 69.276(2.00)
    = 138.6 meters

    We don't know where to go from here...Summer classes. Test every Friday!!! NEED HELP ASAP!
    Thanks
     
  2. jcsd
  3. Jun 8, 2005 #2
    Write down everything the problem gives you.
    F = 5.1 * 10^6 N
    m = ?
    [itex] \Delta [/itex]V = 69.276m/s
    [itex]\Delta[/itex]t = 2s

    What do you think comes next? You're trying to find M.
     
  4. Jun 8, 2005 #3

    Pyrrhus

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    Ok, you know a force made by the catapult produces an acceleration, so the jet will get from 0 to 155 in 2 seconds, you can find the acceleration (throught its definition) and then use Newton's 2nd Law.
     
  5. Jun 8, 2005 #4
    That's too simple. If we were to use F=ma, acceleration would most likely be -9.8 m/s^2. then that would be 5.10 *10^6N/ -9.8m/s^2

    But what is the time used for???
    We were given this equation: X(final) = X(initial) + velocity of x * time + .5(acceleration)(time^2)

    Where we are confused is we were never given an angle that it was catapulted, or a distance to go by, so we are stuck on this next step?!?
     
  6. Jun 8, 2005 #5
    The formula was introduced earlier in the chapter, and it is something we have been using in other problems. I assume that is what we are supposed to be using.
     
  7. Jun 8, 2005 #6

    Pyrrhus

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    You believe the airplane doesn't run a horizontal trajectory before taking off?, This problem, indeed, has several simplifications, i imagine the purpose was what Whozum and i suggested.
     
  8. Jun 8, 2005 #7
    You need to use the impuls equation which I hinted at already. Acceleration would not be 9.8m/s^, it would be 69.276/2 = 34.638m/s^2.

    [tex] F\Delta t = m\Delta v [/tex]

    edit: clarificatoin

    Newton's 2nd Law

    [tex] F = ma [/tex]

    Average acceleration can be expressed as

    [tex] a = \frac{\Delta v}{\Delta t} [/tex] which follows directly from it's definition. Plug that into the above to get what I just gave you.
     
  9. Jun 8, 2005 #8
    Thank you so much! Really appreciate it...don't you hate when its staring you in the face :(
     
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