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Am I right in assuming that if an object, let’s say a car, is travelling with kinetic energy of $x$, then the net work required to stop the object would be $ x $ Joules, but in the opposite direction?

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  • $\begingroup$ A force in the opposite direction does a negative work on the object that brings the kinetic energy to zero. $\endgroup$
    – TaeNyFan
    Dec 9, 2020 at 0:40
  • $\begingroup$ To be clear, what do you (OP) mean by "direction"? Are you thinking of the distinction of work done on or by an object? Or are you thinking of an actual direction that is a property of a vector and not a scalar? $\endgroup$ Dec 9, 2020 at 2:38

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Energy does not have a direction. It is a real number, but it is signed. So the work that must be done on the car will be $W=-x$ but that does not imply a directionality for the work.

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    $\begingroup$ Work is always positive. The minus just helps you keep track of which system is doing the work. $\endgroup$
    – user224659
    Dec 9, 2020 at 0:23
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    $\begingroup$ @TheoreticalMinimum No, the definition of work allows for negative values. In fact, if work is always positive, all potential energies would be negative. Negative work is done by a force when that force has a component opposite the direction of motion in the reference frame being used. $\endgroup$
    – Bill N
    Dec 9, 2020 at 0:53
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    $\begingroup$ @TheoreticalMinimum Work is not always positive. Unless line integrals can't be negative? $\endgroup$ Dec 9, 2020 at 3:22
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    $\begingroup$ @TheoreticalMinimum “you can always give it in positive terms by appropriately chosing the frame of reference” is simply not true. Consider a block sliding across a rough floor. From the floor’s frame the work on the floor is 0 and the work on the block is negative. From the block’s frame the work on the block is 0 and the work on the floor is negative. From the frame of a third object traveling in the same direction but at a slower speed than the block the work on both the block and the floor is negative. Your idea is simply not always correct, and it is rather useless where it is valid $\endgroup$
    – Dale
    Dec 9, 2020 at 12:26
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    $\begingroup$ @TheoreticalMinimum first this is not the place for this discussion. This should be an answer or a chat. Second, what you are describing is not a change of reference frame, it is a change of system. Third, even using change of system it is still not a correct claim. It fails for example in the case of a block sliding across a rough floor and also for a two-body gravitational interaction where the bodies are moving apart. Fourth, the laws of physics work for any system designation, so systems with negative work are valid. Fifth, your idea provides no benefit even where it is not wrong $\endgroup$
    – Dale
    Dec 9, 2020 at 18:24
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Imagine if the car hit a box and instantly transferred all the energy it had to that box. The car would stop, the box would keep going. The kinetic energy the car had is the same as the work it would do on the box hit by it if that object did not resist:

$x=\frac{1}{2} mv^2$

As Dale pointed out, work is a scalar quantity and does not have a direction, but what I suspect you mean is this: "if the box applied just enough force in the opposite direction to keep from moving, so that the car and box were stopped, how much would that work would that force do?"

Well it would indeed be the same as the kinetic energy, yes.

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Yes, that is correct.

Work = Force x Distance

Air and Axle friction would eventually stop it, brakes or going uphill really help. This is why "run away truck" exits on mountain roads are ramps to a higher elevation, using mg(h2-h1) = 1/2mv$^2$ to bring the truck to a halt.

Soft sandy soil is often used to increase drag and help slow the truck as Drag Force x Distance = Work.

The work against the kinetic energy of the truck can be signed "- Work". Since the units of kinetic energy are the same as the units of Force x Distance, we can say, at a halt:

Kinetic Energy - (Sum of all Braking Work) = 0

The rate of Work/Second is referred to as Power.

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