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Kinetic energy ?

  1. Jun 2, 2007 #1
    A net force fo 60N accelerates a 4kg mass over a distance of 10m.
    a) What is the work done by the net force?
    b) What is the increase in kinetic energy of the mass?

    This is what i did, i am not too sure of part b

    F = 60N
    M = 4kg
    d = 10m
    W = Fd
    = 60N*10m
    Answer =600J

    F = ma
    60N = 4kg*a
    a = F/m
    = 15m/s2

    KE = ½(mv2)
    = ½(4kg*15m/s2)
    Answer = 450J
  2. jcsd
  3. Jun 2, 2007 #2
    hey, 15m/s/s is the acceleration not the final velocity.
    you can get the final velocity by V^2 = U^2 + 2*a*d, you take U=0 and V comes out to be sq. root of 300.
    now use KE = (m*v^2)/2 (increase in KE is final ke- initial ke )
    and you ll get 600 J.
    you could have also done it by a very awesome law(hahaha ) called the law of conservation of energy. all the work you are putting in the mass is going to increase the kinetic energy. duh.. so simple
  4. Jun 2, 2007 #3


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    what does work done tell you?

    anyway, in part b) you have worked out acceleration from F... but you need velocity for your KE formula (did u see that?)

    firstly, you are looking for gain in KE, so at least you would have something that looks like
    [tex]\Delta KE = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2[/tex]

    answer is not 450J
  5. Jun 2, 2007 #4


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    Aside: you don't need to take U=0.
  6. Jun 2, 2007 #5
    yes u don't need to take U=0 actually, V ll adjust anyways and you ll still get the increase in KE as 600 joule. sorry for taking a special case.
    also this is a very basic concept, think a bit and you ll yourself figure out the answer.
  7. Jun 2, 2007 #6
    That's the correct calculation of acceleration, but the part you did after that was wrong because you should have used the velocity. Moreover they didn't tell you the initial velocity of this object, so what you really want to get the change in kinetic energy is the change in the velocity squared.

    Now consider for constant acceleration:

    [tex]v_f^2-v_i^2 = 2a\Delta x[/tex]
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