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A 0.50 kg object is at rest. A 3.29 N force

  1. Oct 28, 2003 #1
    Problem 10.
    A 0.50 kg object is at rest. A 3.29 N force to the right acts on the object during a time interval of 1.60 s.
    At the end of this interval, a constant force of 4.07 N to the left is applied for 2.75 s.
    b. What is the velocity at the end of the 2.75 s? Answer in m/s.
    Note: Is the answer -22.385 m/s since in the question it says the force to the left?
     
    Last edited by a moderator: Feb 7, 2013
  2. jcsd
  3. Oct 28, 2003 #2

    HallsofIvy

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    Okay, what have YOU done on this problem so far? What formulas do you think might apply?
     
  4. Oct 28, 2003 #3
    I believe that the impulse-momentum theorem can be used here to slove the problem.
     
  5. Oct 29, 2003 #4

    HallsofIvy

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    F= ma. You know the force and you know the mass so you can calculate the acceleration. Of course, the change in velocity is "acceleration * time".

    To answer your original question, no, the "answer" is not -22.385 m/s. -22.385 m/s is the change in speed due to the second force. What was the speed when the second force was applied?
     
  6. Oct 29, 2003 #5
    Would you add 2.75s to 1.60s to get 4.35s? From that would you then divide 0.50kg from 4.07N, to get 8.14 which is mulitplied by 4.35 to get the answer 35.409 m/s.
     
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