Kinetic Energy Help

Discussion in 'Physics' started by jared11378, Oct 29, 2009.

  1. jared11378

    Thread Starter Member

    Oct 29, 2008
    11
    0
    A 5 * 10^4 kg space probe is traveling at a speed of 11,000 m/s through deep space. Retrorockets are fired along the line of motion to reduce the probe’s speed. The retrorockets generate a force of 4 * 10^5 N over a distance of 2,500 km. What is the final speed of the probe?
    a. 10 000 m/s
    b. 8000 m/s
    c. 6000 m/s
    d. 9000 m/s
    e. 7000 m/s


    I used the equation Wnet = 1/2mv^2(final) - 1/2mv^2(initial)
    Wnet = Fnet d
    Fnet = ma
    Probe = (50000)(11000) = 5.5 * 10^8N
    retrorockets = 4.0 * 10^5N

    Wnet = (5.5 * 10^8N - 4.0 * 10^5)(2500km)
    Wnet = 1.374 * 10^12N

    Solving for V(final) I get

    2(Wnet + 1/2mv^2initial) / m = V(final)^2

    2(1.374 * 10^12 + 3.025 * 10^12) / 5 * 10^4 = Vfinal^2

    I get 13000 m/s and its not a valid answer of course.

    Does anyone see whats going wrong. I was told to use this equation to solve it.
     
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    You plugged in velecity for accelleration. Also you didn't need to bring in F=ma to try and calculate Fnet. Fnet is given to you in the problem.
     
  3. jared11378

    Thread Starter Member

    Oct 29, 2008
    11
    0
    OK I messed up on a few things hear but how does this look.
    d = 2500km = 2.5*10^6m I wasnt converting this to meters

    Wnet = 1/2mv^2(final) - 1/2mv^2(initial)
    Wnet = -Fnet d

    Wnet = (4.0 * 10^5)(2.5*10^6m)
    Wnet = -1.0 * 10^12N

    Solving for V(final) I get

    2(Wnet + 1/2mv^2initial) / m = V(final)^2

    2( -1.0 * 10^12 + 3.025 * 10^12) / 5 * 10^4 = Vfinal^2

    I get 9000 m/s and it looks correct.

    I was thinking Wnet was the vector sum of all forces acting on the object.
    This stuff isnt exactly easy reading. :)

    [​IMG]
     
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