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Homework Help: One force represented along an x-axis

  1. Nov 5, 2008 #1
    1. The problem statement, all variables and given/known data

    The only force acting on a 2.4 kg body as it moves along the x axis varies as shown below. The velocity of the body at x = 0 is 4.0 m/s.

    (a) What is the kinetic energy of the body at x = 3.0 m?

    (b) At what value of x will the body have a kinetic energy of 8.0 J?

    (c) What is the maximum kinetic energy attained by the body between x = 0 and x = 5.0 m?


    2. Relevant equations
    Work is equal to the change in kinetic energy

    3. The attempt at a solution
    I'm at a loss here, I really just need some help interpreting what this graph is. I think a point in the right direction might get me started so any help is appreciated. I've been struggling a bit on work and kinetic energy and were already finishing potential energy in class so any help is appreciated heh
  2. jcsd
  3. Nov 5, 2008 #2
    ok i'm just stabbing in the dark here but this is what i was thinking

    initial Ke + W = Ke final
    so if for part a x=3 and the force is -4 at that point work should be -12.

    the velocity at 0 is 4 m/s so initial kinetic energy is 19.2
    19.2-12=7.2 kinetic energy at x=3
  4. Nov 5, 2008 #3
    ok i figured it out on my own and for anyone else wondering out there about this problem lemme explain.

    using what i said in the before post
    initial Ke + W = Ke final (which i'll call Kei and Kef now)
    at x=3 the work area added up under the line is 2 (at 0) 0 (at 1) -2 (at 2) and -4 (at 3). Add up those areas for a total area which gives you the work of -4. Plug that W into Kei + W = Kef, and since we already know Kei from whats givin in the problem we get the answer.

    part b is working backwords using the same kei + an unknown W = 8.
    what you get for work is 11.2 and with 8 W at 4 and 12 W at 5 it's just a matter finding where in between you need to be.

    you should be able to get C if you got this far but if not lemme know and i'll help more
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