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Wikipedia's defintion of the joule is confusing me

  1. May 24, 2010 #1
    According to Wikipedia, a joule is "the energy exerted by the force of one newton acting to move an object through a distance of one metre." That sounds like a joule is very inconsistent. We could apply one newton to a bowling ball to make it move one meter, and we could apply one newton to a marble to make it move a meter. I think the marble will require far less time and energy.

    I was going to complain about the mathematical definition of the joule too, but, now that I've rejected Wikipedia's definition, and tried fruitlessly to simplify the meter out of the formula for the joule, I think I understand it now. I think that a joule is better described as the force required to accelerate a 1 kg object at 1 m/s/s for one second, or, to simplify that, a joule is the energy required to increase a 1 kg object's velocity by 1 m/s. Is this correct?
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  3. May 24, 2010 #2


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    Time does not have any factor in terms of energy in the way you are thinking. Work is expended by applying a force over a given distance (in the mechanical sense although it has been shown that other ideas of work in terms of thermodynamics, electromagnetics, etc are equivalent). Sure it will take far less time to do the same amount of work on a marble than a bowling ball for a given force, but this sense of time is not part of the definition (that would be power you are thinking of which is the amount of work expended over a given interval of time).

    As for the amount of energy to change an object's velocity? No. The change in a velocity is the acceleration. We can accelerate an object without doing any work because work is the dot product of force and displacement. If we apply a force that is normal to the path of displacement, no work is done despite the fact that the object undergoes an acceleration. For example, a stable planetary orbit of a planet around a star is a situation where the planet is undergoing constant acceleration but no work is done. True, in these cases the component of the force that is not doing any work is only affecting a change in the direction of the velocity and not a change in its magnitude.

    So, look at kinetic energy which is 0.5*m*v^2. This should show you that a change in velocity is related to the square root of the energy. The amount of energy expended to increase an object's speed by 1 m/s is:
    [tex] \Delta E = 0.5m(v+1)^2-0.5mv^2 = mv+0.5v^2 [/tex]
    Thus, the amount of energy expended was dependent upon the mass and initial velocity of the object. So defining a Joule as the amount of work expended to accelerate a mass of 1 kg by 1 m/s only works if the mass was initially at rest. But that's not all, we only looked at the kinetic energy. There is a possible potential energy too. The amount of work that you expended to move that mass up a hill to get it going 1 m/s is different than on a flat plane. So this definition is not sufficient.
  4. May 24, 2010 #3
    There's a part missing from that definition. Everything else is correct, but as you pointed out, mass must be given. A joule is the energy given by a force exerted on a 1 kg object displaced 1 meter.
  5. May 24, 2010 #4


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    The mass is incorporated in the force. The amount of work done is given by:
    [tex] W = \int_a^b \mathbf{F} \cdot d\mathbf{\ell} [/tex]

    The mass of the object is only a factor in terms of the force. Stating that the 1 Joule is the amount of work done by applying a 1 N force over a distance of 1 m is acceptable I believe.
    Last edited: May 24, 2010
  6. May 24, 2010 #5
    When you have learnt about work and kinetic energy theorem,you will understand it.
  7. May 24, 2010 #6


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    Well, the reason that the mass of the object doesn't have to be specified is that it drops out of the equation.

    Here's the rub: the heavier the object the slower the final velocity. And there is an exact relation between the mass of the object and the velocity it reaches.

    I define the following variables:

    s distance covered, wich in this case is 1 meter
    v velocity
    a acceleration

    t time
    m mass of the object
    F the force upon the object, which in this case is 1 newton

    [tex]F=ma[/tex] => [tex]m = 1/a[/tex]

    [tex]v = at[/tex] => [tex]t=v/a[/tex]
    [tex]s = \frac{1}{2} a t^2[/tex]

    Substituting for 't' to get rid of the variable 't', and setting 's' to 1 meter:

    [tex]1 = \frac{1}{2} \frac{v^2}{a}[/tex]

    [tex]1 = \frac{1}{2} m v^2[/tex]

    What this shows is that whatever the mass 'm' is, if the force is 1 newton, and the distance over which it acts is 1 meter, then the final velocity is dependent on the mass; the bigger the mass the slower the final velocity

    In the final state the expression [tex]\frac{1}{2} m v^2[/tex] will always evaluate to 1 unit. We call this quantity 1 Joule.
  8. May 24, 2010 #7


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    So mass has an impact on acceleration and time but does not affect the work.
  9. May 24, 2010 #8

    Meir Achuz

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    This statement is wrong. The mass is irrelevant. Work= F X D.
  10. May 25, 2010 #9

    Andrew Mason

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    This is not correct. The increase in energy of a mass in going from v to v+dv is:

    [tex]\Delta KE = \frac{1}{2}m(v+dv)^2 - \frac{1}{2}m(v)^2 = \frac{1}{2}m(v^2 + 2vdv + dv^2 - v^2) = mvdv + \frac{1}{2}mdv^2[/tex]

    You can see from this that the change in energy is not only a function of the change in speed, but of the initial speed (v).

    It requires much more energy to change the speed of the 1kg mass from 10m/s to 11 m/s than it does from 0 to 1 m/s (10.5 J compared to .5 J)

  11. May 25, 2010 #10
    So then energy required to cause two objects to move away from each other increases exponentially as velocity increases. These two objects might be things like a cannon and cannonball, or a rocket and rocket fuel, right?

    And, if we take a rocket as an example again, it doesn't matter if both the rocket and fuel are moving relative to Earth, or some other point, it only matters how fast the fuel moves away from the rocket, right?

    For one final example, if I'm sitting on a 747 going nearly the speed of sound, and I want to push a marble, say, 10 cm on my tray, it won't take exponentially more energy than doing the same thing at home, right?

    Finally, how do I go about calculating these energy requirements quickly? The only way I can think of to do it is to pick some arbitrary length and then iterate an object over that length again and again while tracking velocity. This seems like I'll have to choose between making the calculation very slow, or making it very inaccurate.
  12. May 26, 2010 #11

    Andrew Mason

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    The force on the rocket is the time rate of change of momentum of the expelled gases. So, yes, for a given mass being expelled, the change in speed of the rocket is proportional to the speed of the expelled mass.

    You have to be careful with rockets and energy, though. The energy of the rocket engine results in kinetic energy being imparted to the expelled gases and to the rocket. If you measure kinetic energy in the earth frame, the speed of the expelled gases relative to the earth changes with rocket speed. The speed of the centre of mass of the rocket+gases cannot change (once the rocket is in empty space).

    Pushing the marble forward on the plane requires an equal and opposite (ie. backward) push on the plane. So, the motion of the centre of mass of the plane itself does not change; hence the kinetic energy of the plane + marble does not change.

    Calculus would be needed. The force on the rocket is F = dp/dt = vdm/dt where v is the nozzle speed of the expelled gas and dm/dt is the rate of mass flow of the expelled gas. It is complicated further by the fact that the mass of the rocket is decreasing as fuel is burned and gases expelled.

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