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Conservation of energy (Finding Velocity)

  1. Jun 1, 2013 #1
    1. The problem statement, all variables and given/known data

    The 100kg crate shown in Fig. 1 is acted upon by a force having a variable magnitude
    P = 20t N, where t is in seconds. Determine the crates’ velocity 2 seconds after P has
    been applied. The initial velocity is v1=3m/s down the plane, and the coefficient of
    kinetic friction between the crate and the plane is μk = 0.3. (60 points)
    Hint) For easy calculation, the gravitational acceleration (g) is assumed 10m/s2
    2. Relevant equations
    T1 + U1-2 = T2
    T kinetic energy
    U potential Energy
    3. The attempt at a solution
    I am having problem finding the distance S ???
    IMG_0871.jpg
     
  2. jcsd
  3. Jun 1, 2013 #2

    Doc Al

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    Staff: Mentor

    Think in terms of momentum instead of energy.
     
  4. Jun 1, 2013 #3
    Aha ....
    mv1 + P t (dlta t) + mg sin 30 (dlta t) - mg cos 30 Mk (dlta t) = mv2
    30 + 80 + 1000 + 519.6 = 100 v2
    v2 = 5.9 m/s
     
  5. Jun 1, 2013 #4

    Doc Al

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    Staff: Mentor

    Double check the first two terms.
     
  6. Jun 1, 2013 #5
    300 + 80 + 1000 - 519.6 = 100 v2
    v2 = 8.6 m/s
    Shouldn't the friction be negative and what about the P , It is given P = 20 t , shouldn't i multiply it by t=2 twice ??
     
  7. Jun 1, 2013 #6

    Doc Al

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    Staff: Mentor

    Yes.

    Since the force P is a function of time, you'll need to integrate.
     
  8. Jun 1, 2013 #7
    I was wrong about P . P = 20 t → Impulse = ∫ p dt = 20 t2 /2
    Thus P = 20 (4) / 2 = 40 N

    But isn't the friction impulse negative and thus we get

    300 + 40 + 1000 - 519.6 = 100 v2
    v2 =8.2 m/s ????
     
  9. Jun 1, 2013 #8

    Doc Al

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    Staff: Mentor

    Good!
     
  10. Jun 1, 2013 #9
    Thanks ....
     
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