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  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Imagine a black box8, containing a gasoline-powered engine, which is designed to reel in a steel cable of length d, exerting a certain force F.

    If we use this black box was to lift a weight, then by the time it has pulled in its whole cable, it will have lifted the weight through a height d. The force F is barely capable of lifting a weight m if F=mg, and if it does this, then the upward force from the cable exactly cancels the downward force of gravity, so the weight will rise at constant speed, without changing its kinetic energy. Only gravitational energy is transferred into the weight, and the amount of gravitational energy is mgd, which equals Fd. By conservation of energy, this must also be the amount of energy lost from the chemical energy of the gasoline inside the box.9

    Now what if we use the black box to pull a plow? The energy increase in the outside world is of a different type than before: mainly heat created by friction between the dirt and the ploughshare. The box, however, only communicates with the outside world via the hole through which its cable passes. The amount of chemical energy lost by the gasoline can therefore only depend on F and d, so again the amount of energy transferred must equal Fd.

    The same reasoning can in fact be applied no matter what the cable is being used to do. There must always be a transfer of energy from the box to the outside world that is equal to Fd. In general, when energy is transferred, we refer to the amount of energy transferred as work, W. If, as in the example of the black box, the motion of the object to which the force is applied is in the same direction as the force, then W=Fd.


    x / The baseball pitcher put kinetic energy into the ball, so he did work on it. To do the greatest possible amount of work, he applied the greatest possible force over the greatest possible distance.

    If the motion is in the opposite direction compared to the force, then W=-Fd; the negative work is to be interpreted as energy removed from the object to which the force was applied. For example, if Superman gets in front of an oncoming freight train, and brings it to a stop, he's decreased its energy rather than increasing it. In a normal gasoline-powered car, stepping on the brakes takes away the car's kinetic energy (doing negative work on it), and turns it into heat in the brake shoes. In an electric or hybrid-electric car, the car's kinetic energy is transformed back into electrical energy to be used again.


    w / The black box does work by reeling in its cable.

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