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Law of conservation of energy

  1. Dec 30, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle of mass 0.400 kg is shot from P . The particle has an initial velocity vi with the horizontal component of 30 m/s. The particle rises to a maximum height of 20 m above P. Using the law of conservation of energy, determine :
    (a) the vertical component of vi
    (b) the work done by the gravitational force on the particle during its motion from P to B
    (c) the horizontal and the vertical components of the velocity vector when the particle reaches B
    P is in height of 60 m.

    2. Relevant equations
    Ek = -Ep

    3. The attempt at a solution
    a.)
    1/2mv2 = -(m*g*h)
    1/2v2 = -(-9,8*20)
    v2 = 2*9,8m/s2*20m
    v=19,7 m/s¨
    b.)
    A=m*g*h
    A=0,4kg*9,8m/s2*60m
    A=294 J
    c.)
    1/2mv2 =-(m*g*h)
    1/2v2=-(9,8*-80)
    v2=2*9,8m/s2*80m
    v=39,5m/s

    I do not know if I should calculate with minus Ep. And then if the acceleration should be with minus in a.) and in c.) the acceleration positive and height negative?
     
  2. jcsd
  3. Dec 30, 2012 #2
    I am a bit confused. You say the particle reaches a max heighth of 20 m but use use 60 m in the solution. Seems like part a is correct. The first statement of part b is correct but the second line makes no sense to me.
     
  4. Dec 30, 2012 #3
    Sorry, it is confusing. The particle is shot from P (P is in the height of 60m above the ground) and then particle rises to height of 20m above P so the particle is in the total height of 80m above the ground and then reaches the ground - B . I hope that I made it clear.
     
  5. Dec 30, 2012 #4

    haruspex

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    You can choose to measure distance upwards or downwards - entirely up to you, so long as you are consistent. Note that swapping swaps the sign on both acceleration and height, so leaves the sign of PE unchanged.
     
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