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Work done on free falling object as function of time

  1. Mar 22, 2013 #1
    Hi there everyone,

    Would like some checking of my work or comments.
    Would like the work done on a falling mass M as a function of time.
    To begin, the relation for work as function of distance..

    w = Mgx

    but would like as function of time so substitute x this way,

    x = (1/2)gt2

    implies work done as function of time would be

    w = Mg(1/2)gt2

    and finally for work as function of time;

    w = (1/2)Mg2t2

    Thanks in advance for any comments, verifications,
    or kibbutzing in general!
  2. jcsd
  3. Mar 22, 2013 #2

    Doc Al

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    That's fine. That's the work done by gravity on a falling mass (starting from rest) after a time t.
  4. Mar 22, 2013 #3
    Thanks there Doc for your kind and swift reply!
  5. Mar 22, 2013 #4


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    A simpler way of stating it would be mgh, I think.
  6. Mar 22, 2013 #5

    Doc Al

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    That was the starting point, his first equation. But he wanted a function of time.
  7. Mar 22, 2013 #6


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    That'll teach me to read the whole thread.
    Interestingly, the power increases with time.
  8. Mar 23, 2013 #7
    Thanks SophieC for comments,
    yep, that is interesting, ..if power
    is defined as energy delivered within
    a time period t, ie,

    p = w/t

    we'd get in that outlook,

    p = (1/2)Mg2t,

    which is linearly proportional to the time.

    However, if I'm not mistaken, power I think might be supposed
    to be more of an instantaneous concept, i.e.,
    the small change in work done over a short time period;

    p = Δw / Δt

    in which case the problem might start to require
    a bit of a calculus based treatment.
  9. Mar 23, 2013 #8

    Doc Al

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    Exactly right. What you've found is the average power over that time period. To find the instantaneous power, take the derivative: p = dw/dt.

    Why don't you give it a shot?
  10. Mar 23, 2013 #9
    well, i'll give that a shot,...

    I 'd say that in a constant gravitational field, we know

    w = mgx

    But mg = const, so differentiating both sides ...

    dw = mg dx

    but since

    x = (1/2)gt2

    we differentiate each side to get;

    dx = (1/2)g2t dt = gt dt

    so that

    dw = mg gt dt = mg2t dt

    and dividing by dt on both sides,..

    dw/dt = mg2t

    but power is change in work within time period, so
    the power is,.

    P = mg2t

    which is a weird result because the power delivered
    to the system increases with time, whereas there is
    still constant acceleration. It would be like if you had
    a rocket with an engine that, even though it puts out a constant force,
    the power it delivers keeps increasing.

    We could check that result in another way,..
    we can write that the work is the force times the distance,
    written differentially with constant force,..

    dw = F dx

    and we could divide both sides by dt,..

    dw/dt = F (dx/dt)

    which reads...

    P = Fv

    where v is the velocity dx/dt.

    which shows the same result, that the power
    delivered increases with the velocity or time,
    which would seem counter-intuitive a bit
    since it appears you can't apply any power
    to a stationary object,.. but we know that
    to get your car moving the engine must be
    running and engaged in delivering power and using gas.

    Maybe it could be reckoned with by considering
    your car or rocket tied to the ground with chains,
    and if it can't move at all then you can apply
    all the engine force you want but like it or not
    you are not delivering any power to the car.

    An interesting philosophical point is that if you're
    dealing with an instantaneous new occurrence,
    like a bomb going off you could employ the whole science of
    statics instead of dynamics,... that is
    sum of all force vectors = zero
    ƩF = 0.
    Since all velocities equal zero initially.
    That is, you could introduce the forces of
    the bomb into the building structure original blueprints, and using
    the usual engineering science of statics you could
    calculate all stresses throughout the building to see
    where failure might occur, i.e., which stresses would be
    enough for some failure,... i.e., acceleration of some part
    of the building,.. in which case it would exit the science
    of statics and enter the science of dynamics.

    Sorry if I've rambled on a bit, it is late and just thinking out loud.
  11. Mar 24, 2013 #10
    Wait a second ... you're trading potential energy for kinetic energy. The total energy remains the same, hence no work is performed.
  12. Mar 24, 2013 #11
    Work = force x distance(basically)
    Work is done on the falling mass by the force of gravity. This work done converts potential energy to kinetic energy (+ any energy due to air turbulence/friction)
  13. Mar 24, 2013 #12


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    Difficult to avoid when you stray into Science!
    There is no special significance to this speed - power thing except that we tend to think in terms of cars etc, which have a specific max power output and, hence, a top speed, defined by when engine power is balanced by 'speed times friction forces'. I often find it interesting just to turn things around and look at them slightly differently - producing counter-intuitive conclusions.
    Of course, the process of falling under gravity has a limit too. When you hit the ground.
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