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Homework Help: Force given mass, velocity, and displacement

  1. Jun 16, 2010 #1
    1. The problem statement, all variables and given/known data
    A 6.0 g bullet leaves the muzzle of a rifle with a speed of 338 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.9 m long barrel of the rifle?


    2. Relevant equations
    F=ma
    x=vt
    a=v/t

    3. The attempt at a solution
    x=vt
    .9m=338m/s t
    t=.0027s

    a=v/t
    a=338 m/s /.0027s
    a=126937.8 m/s2

    F=ma
    F=.006kg (126937.8 m/s2)
    F=761.6N
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 16, 2010 #2

    kuruman

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    You cannot use x = vt. It is valid only when the acceleration is zero - not the case here. Instead, start with the kinematic equation that relates displacement, acceleration and speed and has no explicit time dependence.
     
  4. Jun 16, 2010 #3
    First let me just ask you a question. Is your context now up to energy (or example, conservation of energy)?

    For this problem you are giving the length of the path, or the height.
    And you have the final speed.

    Using conservation of energy you will be able to find the the initial speed. This will allow you to compute the force using work-kinetic energy theorem which involves velocity (in this case our initial velocity).

    However, using your approach. It can be done too.

    What you need didn't take into account is exactly the notion of "initial speed".
    Have your initial speed compute, and use whatever you have, find the acceleration, and finally get the force.
     
  5. Jun 16, 2010 #4
    v^2=Vo^2 + a x is this right?
    338^2 m/s=a (.9m)
    a=63468.9 m/s^2

    F=ma
    F= .006kg (63468.9m/s^2)
    F=380.8N

    correct?
     
  6. Jun 16, 2010 #5

    collinsmark

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    Sorry, but the above 'x = vt' equation only applies to objects moving at a constant velocity. This problem assumes a constant force (thus constant acceleration) which is quite different. Your value for t isn't correct.

    If you must use kinematics to solve this problem, try one of your other equations that assume a constant acceleration. (It is possible to solve this problem using kinematics but there is an easier way [see below]).

    But I should point out that there is an easier way to solve this problem using the work-energy theorem; a way of expressing conservation of energy. Here are some questions,
    (a) What is the definition of work, W, done by a constant force over a given distance?
    (b) What is the definition of kinetic energy, given an object's mass and velocity?
     
  7. Jun 16, 2010 #6

    collinsmark

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    You left out a '2', but you got the right answer below so I assume it was just a typo. Just to be clear, the equation is,

    vf2 = v02 + 2ax

    Looks okay to me :approve:

    (ps. if you are up to the point in your coursework where are are studying work and conservation of energy, I suggest you solve it that way too. If not, you have something to look forward to.)
     
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