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Proving the Work-Energy Theorem A lot of calculus, anyone?

  1. Nov 17, 2011 #1

    A particle moves on a curved path from (x1, y1,z1) to (x2, y2,z2). At the start, the particle has a velocity of v = v1xi+v1yj+v1zk. This curved path can be divided into segments infinitesimally, which are, dl = dxi +dyj +dzk. It is acted on by a net force F = Fxj + Fyi + Fzk. The force components Fx, Fy, and Fz, are in general functions of position. Prove the work energy theorem for this general case. That is prove that



    Wtot=[itex]\int[/itex]F dl = [itex]\int[/itex] Fxdx + Fydy + Fzdz

    where the limits of integration are from (x1, y1,z1) to (x2, y2,z2) for each.

    2. Relevant equations

    W = Fd

    ax= [itex]\stackrel{dvx}{dt}[/itex] = [itex]\stackrel{dvx}{dx}[/itex] *[itex]\stackrel{dx}{dt}[/itex] = vx[itex]\stackrel{dvx}{dx}[/itex]

    3. The attempt at a solution

    I don't really have any idea where to start but what I did was I took the velocities and made them into accelerations then changed the Fx, Fy, and Fz, into max, may, and maz values, then I'm confused now because I have an integral which looks like this..

    [itex]\int[/itex] m*vx*dvx + m*vy*dvy/dx dy + m*vz*dvz/dx dz

    which makes no sense because the dx only cancelled out for the X direction, and I need to prove that Wtot = 1/2 mvf^2 - 1/2mvi^2

    If anyone could help it'd be appreciated because i've got really no idea what im doing
  2. jcsd
  3. Nov 17, 2011 #2
    Bolded means vectors. Note that x = (x1,x2,...xn). The period denotes dot product.

    Let W = ∫F.dx
    F = m(d2x/dt2)

    W = m∫(d2x/dt2).dx
    W = m∫(dv/dt).dx

    Note that this step is somewhat hand-wavy but it's true. I think it would be a good exercise for you to prove this. To do that, you first need to learn about how line integrals work, and the process of parametrization. Hint: dx=(dx/dt)dt

    W = m∫dv.(dx/dt)
    W = m∫(dx/dt).dv
    W = m∫v.dv
    W = m(1/2 v.v) = 1/2 m|v|2

    Is your name Ping?
    Last edited: Nov 17, 2011

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