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Homework Help: Solve for acceleration

  1. Oct 4, 2007 #1
    1. The problem statement, all variables and given/known data

    A freight train has a mass of 1.20E+7 kg. If the locomotive can exert a constant pull of 6.26E+5 N, how long does it take to increase the speed of the train from rest to 82.4 km/hr?

    What does the 1.20E + 7 kg mean? What is E?

  2. jcsd
  3. Oct 4, 2007 #2


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    1.20E + 7 means [tex]1.20 \times 10^7[/tex]

    E means exponent.
  4. Oct 4, 2007 #3
    All right I'm really not understanding these forces...

    So the mass is = 1.2E + 7 kg and the pull from the Locomotive is 6.26E+5 N

    so I thought I would want to solve for acceleration and then figure out how long it takes to get to the velocity of 82.4 km/h

    So I said that F = ma
    so 6.26E+5 = 1.2E+7 (a)
    so a = 0.052 m/s

    So then v=v0 + at
    82.4 = 0.052t
    1584.6 = t

    Which i don't think is anywhere's near right at all....What am I missing? I think i'm probably missing some other forces...
  5. Oct 4, 2007 #4


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    You did everything right except you didn't convert 82.4km/hr to m/s.
  6. Oct 4, 2007 #5
    Hey thanks!
  7. Oct 4, 2007 #6
    All right I've got another question for you :D

    A 5.26g bullet leaves the muzzle of a rifle with a speed of 328 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.792 m long barrel of the rifle?

    Ok, so I know my initial velocity is 328 m/s. I'm not sure of the significance of the .792m long barrel, and I'm not sure of what forces are acting on the bullet... I was thinking that you would need to solve for the acceleration in order to use F=ma...am I on the right track?
  8. Oct 4, 2007 #7


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    Yes, you need acceleration. But the path over which we are getting the acceleration is over the barell of the rifle... so 328m/s is the final velocity not the initial velocity.
  9. Oct 4, 2007 #8
    Does that mean that my inital velocity is 0?
  10. Oct 4, 2007 #9


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  11. Oct 4, 2007 #10
    ok so
    delta x = 1/2 (v initial + v final)t
    .792 m = 1/2 (328) t
    .792 m = 164 t
    0.0048 = t (Is that right?)

    Then to find the acceleration

    v=v initial + at
    328 = a(.0048)
    68333 = a ??
  12. Oct 4, 2007 #11


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    yeah, that's right, but watch the rounding... I'm getting 67,919m/s^2.

    You can also use the equation:

    vf^2 = vi^2 + 2ad

    to get acceleration.
  13. Oct 4, 2007 #12
    You are definiately my hero, you know that ;)

    I've only got one more question for tonite I promise *Grins*

    Two blocks of mass m = 12.6 kg each are fastened to the ceiling of an elevator, as seen in the figure below. (well obviously you can't see them)
    Anyways, the blocks are hanging one directly below the other, so block A is suspended to the ceiling of the elevator, and block B is suspended to the bottom of Block A

    The elevator accelerates upward at a = 2.33 m/s2. Find the tension in the bottom rope.

    Well good news, I actually did this part *Cheers* The answer is 153 N.

    However, I'm not sure how to approach the second part of the problem:
    Find the tension in the top rope.

    I assume I use the answer I found for the first part, but I'm not sure how

    I thought it would just be
    Sum of forces=ma
    Ft = 182.358 N

    But as is usual, this isn't the right answer ;) Can you spot where I'm going wrong?
  14. Oct 5, 2007 #13


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    You are very close. For the sum of forces, you forgot to take into account gravity... So you have Ft, the 153N downward force from the bottom rope, but also the weight of the block.
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