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Work/energy related

  1. Dec 5, 2003 #1
    A time-varying net force acting on a 5.8kg particle causes the object to have a displacement given by

    x = a + bt + dt^2 + et^3

    where a = 2m, b = 1.8 m/s, d = -1.6 m/s^2, e = .94 m/s^3 with x in meteres and t in seconds. Find the work done on the partciel in the first 4.4 seconds of motion. Answer in units of J.

    i've taken both derivative and integral, im not getting any way correct grr. can someone help
     
  2. jcsd
  3. Dec 6, 2003 #2
    ...

    Heres how u do the problem...


    [tex]\int dw = \int_{x_{1}}^{x_{2}} F.dx[/tex]; where [tex]F[/tex] is the force and is equal to [tex] m*a[/tex], where [tex]m[/tex] is the mass and [tex] a [/tex] is the accln.
    To find the work done by the force in 1st 4.4 seconds you will have to do this:

    [tex] x = a + bt + dt^2 + et^3 [/tex]. Therefore:

    [tex] dx = (b + 2dt + 3et^2)dt [/tex]

    substitute this in the work equation to get:

    [tex] \int dw = \int_{0}^{4.4} F * (b + 2dt + 3et^2)dt [/tex]

    where:
    [tex]F = m*a[/tex], and [tex] a = d^2x [/tex] for the 1st 4.4 seconds. Find this force and then substitute in the work integral..


    There is another way to this problem:
    work done = change in kinetic energy. Find the kinetic energy at t = 0 and at t = 4.4 by finding the velocities at t = 0 and at t = 4.4sec by differentiating the expression for x and then substituting for t.

    [tex]work done = W = m(v_{4.4}^2 - v_{0}^2)/2 [/tex]....

    Sridhar
     
    Last edited: Dec 6, 2003
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