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Math Help - Conservation of Energy

  1. #1
    Super Member craig's Avatar
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    Conservation of Energy

    Taking the value of g to be 9.8 ms^{-2}.

    Body of mass 5kg falls freely from rest a distance of 3m, find the kinetic energy acquired by this body.
    Let the height above the ground be h.

    Initially KE = 0, and Gravitational Energy (PE) = mgh = 5gh.

    Due to the conservation of energy, we know that this total of energy stays the same. So at our point 3 metres below the starting point:

    KE = 2.5v^2, PE = 5g(h-3).

    Not sure how to work out the KE acquired by the body, and have I written the equations right?

    Part 2: If the particle is brought to rest by a force of 70N, find the further distance fallen.
    No idea where to start on this one, any help would be greatly appreciated.

    Thanks again in advance.
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    Body of mass 5kg falls freely from rest a distance of 3m, find the kinetic energy acquired by this body.

    \Delta KE = -\Delta GPE

    \Delta KE = -(mgh_f - mgh_0) = -mg(\Delta h) = -mg(-3) = 3mg<br />

    go ahead and find the velocity ... you'll need it for part 2.

    Part 2: If the particle is brought to rest by a force of 70N, find the further distance fallen.

    F_{net} = ma

    70 - mg = ma

    a = \frac{70 - 5g}{5}

    v_f = 0

    now use the equation v_f^2 = v_0^2 + 2a(\Delta y) to determine \Delta y ... remember that \Delta y < 0
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    Super Member craig's Avatar
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    Quote Originally Posted by skeeter View Post
    Body of mass 5kg falls freely from rest a distance of 3m, find the kinetic energy acquired by this body.

    \Delta KE = -\Delta GPE

    \Delta KE = -(mgh_f - mgh_0) = -mg(\Delta h) = -mg(-3) = 3mg<br />

    go ahead and find the velocity ... you'll need it for part 2.
    Thanks for the reply.

    For the velocity, \frac{1}{2}mv^2 = 3mg., so v = \sqrt{6g} \approx 7.668

    Just to clarify, h_f is the initial height?

    Quote Originally Posted by skeeter View Post
    Part 2: If the particle is brought to rest by a force of 70N, find the further distance fallen.

    F_{net} = ma

    70 - mg = ma

    a = \frac{70 - 5g}{5}

    v_f = 0

    now use the equation v_f^2 = v_0^2 + 2a(\Delta y) to determine \Delta y ... remember that \Delta y < 0
    So from your equation, v_f^2 = v_0^2 + 2a(\Delta y).

    So 6g + 2(14 - g)(\Delta y) = 0

    \Delta y = \frac{-6g}{2(14 - g)} = \frac{-3g}{14 - g} = -7

    So the further distance fallen is 7 metres?
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    Quote Originally Posted by craig View Post
    Thanks for the reply.

    For the velocity, \frac{1}{2}mv^2 = 3mg., so v = \sqrt{6g} \approx 7.668

    Just to clarify, h_f is the initial height?

    no, \textcolor{red}{h_f} is final height



    So from your equation, v_f^2 = v_0^2 + 2a(\Delta y).

    So 6g + 2(14 - g)(\Delta y) = 0

    \Delta y = \frac{-6g}{2(14 - g)} = \frac{-3g}{14 - g} = -7

    So the further distance fallen is 7 metres?

    correct
    ...
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    Super Member craig's Avatar
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    Thanks again.
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