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Force on a particle

  1. Jul 22, 2013 #1
    A 500g particle has velocity vx = -4.0m/s at t = - 2 s. Force Fx = (4−(t/s)2) N is exerted on the particle between t = - 2 s and t = 2 s. This force increases from 0 N at t = - 2 s to 4 N at t = 0 s and then back to 0 N at t = 2 s. What is the particle’s velocity at t = 2 s?
    I'm not sure if the answer is 4 m/s. can someone check if I do it right?

    V2 = V02 + at
    v= v0+ (F/m)t
    v = -4m/s + (0/0.5)*2
    v= 4m/s
     
  2. jcsd
  3. Jul 22, 2013 #2

    mfb

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    Staff: Mentor

    v= v0+ (F/m)t requires a constant force, or you have to calculate the average force. You cannot use the force at some arbitrary point in time here.

    The first equation looks odd.
    How did you get the last line from the line before?
     
  4. Jul 22, 2013 #3
    It doesn't have constant force so which formula do I use?

    v = 4 m/s?
    I got that from plug in v02
    v0 = -4 m/s
    so I thought it would be 4 m/s since I have to take square root to get v
    Is it wrong??? lol
     
  5. Jul 23, 2013 #4

    mfb

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    How are the velocity as function of time and the force related?

    Alternatively, calculate the average force.

    The first formula was wrong anyway. You can see it if you check the units.

    v2=16m2/s2 has two solutions: v=4m/s and v=-4m/s. But the formula you used was not right anyway.
     
  6. Jul 24, 2013 #5
    I figure it out!!! Thank you :)
     
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