The Wikipedia definition of Joule is:

the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of its motion through a distance of one metre (1 newton metre or N⋅m).

How does it make sense to define it in terms "through a distance of one metre"? Isn't that relative to the object's mass and already existing velocity, making the definition of Joule variable?

Also I see a lot of examples that give lifting an object that weights 1 newton (like an apple, or a tomato in Wikipedia's example) 1 meter, but wouldn't applying 1N to an object that weights 1N means the forces will cancel out and it wouldn't move anywhere, just hover in place?

  • $\begingroup$ The force must equal the resistance. It could be a small mass on a rough surface or a big mass on wheels. $\endgroup$ – paparazzo Jan 24 '18 at 12:14

It's true that after experiencing a force of 1 Newton through 1 meter of distance, objects of different masses and initial velocities will have different ending velocities. But, we are dealing with energy, which takes these into account. Let's see what happens when an object with unknown mass and unknown initial velocity experiences a constant force.

From $F=ma$, we known that the acceleration due to a constant force will be constant. Under constant acceleration, an object will travel a distance given by $$x = \frac{1}{2}at^2 + v_0t$$ where $x$ is the distance traveled, $a$ is the induced acceleration, $t$ is the travel time, and $v_0$ is the initial velocity. Since acceleration is constant, we can write $$a = \frac{v_1 - v_0}{t}$$ where $v_1$ is the object's final velocity. Substituting: $$x = \frac{1}{2}\left(\frac{v_1 - v_0}{t}\right)t^2 + v_0t$$ $$x = \frac{v_1 + v_0}{2}t$$ Multiply both sides by $ma$: $$max = \frac{v_1 + v_0}{2}mat$$ Using $F=ma$ on the left and $at = v_1 - v_0$ (second equation above) on the right: $$Fx = \frac{1}{2}m\left(v_1 + v_0\right)\left(v_1 - v_0\right)$$ $$Fx = \frac{1}{2}mv_1^2 - \frac{1}{2}mv_0^2$$ Since kinetic energy is defined as $K = \frac{1}{2}mv^2$, we can write the above as $$Fx = K_1 - K_0$$ Applying a force over a distance results in a well-defined change in energy, no matter the object's mass or velocity. Since energy is a useful value to calculate, we assign a name to that which causes changes in energy, namely work: $$W = Fx$$

  • $\begingroup$ Why t disappeared after you added in ma? $\endgroup$ – yoni0505 Jan 24 '18 at 11:30
  • $\begingroup$ @yoni0505 I used the equation for acceleration to substitute. I've edited my answer. $\endgroup$ – Mark H Jan 24 '18 at 11:35

The distance moved is unambiguous. When we say "John ran 100 meters in 15 seconds", we don't need to worry about what John's mass is. (The more common language here is "body weight", but this is a physics forum). We also don't need to worry what his starting velocity was. Of course, if John started from a running speed, we would expect him to finish the 100m run faster, but that's all. In the same way, statements such as "the distance between Chicago and Toronto is 702 kilometers" are unambiguous.

If you push a box with a $1N$ force in a straight line for $1m$, then you've done one joule of work. It's irrelevant how massive the box is or how fast it was originally moving. Both of these things affect the box's original kinetic energy, but not the amount of kinetic energy you impart to the box.

  • $\begingroup$ but the amount of energy required to push that box 1m depends on its mass and velocity, so that would mean by this definition that 1 joule could represent any amount of energy. If joule was defined in terms of force and time (instead of distance) that would make sense, since time doesn't depend on the object's properties. For example this definition makes sense: "It is also the energy dissipated as heat when an electric current of one ampere passes through a resistance of one ohm for one second". $\endgroup$ – yoni0505 Jan 24 '18 at 10:49
  • $\begingroup$ @yoni0505 The amount of energy required to push that box 1m depends on its mass and velocity, yes. If the box is already moving you don't even need to push it and it'll move 1m. But if you want to impart 1J of energy to the box, you'll need to use 1N of force for the 1m distance (more precisely, you'll need the integral of F dS to be 1). $\endgroup$ – Allure Jan 24 '18 at 11:00

You have to consider what "Energy transfer" (Work) is. Work is how much force is applied on an object over a certain distance.

Given this, a Joule (the unit of Energy and Work) is equal to a Newton (unit of force) * a meter (unit of distance).

If a box has had 10 Joules of work applied to it and has moved 5 meters, this tells you that 2 Newtons of force must have been applied.

This is all work/energy transfer is: force applied over a certain distance. Given this, it makes sense that the unit of work (a joule) is equal to the unit of force (a newton) times the unit of distance (a meter).

  • $\begingroup$ but the distance it moves depends on the time and environment. if I apply those 2N on an object in perfect space, it would move infinite amount of distance. How do you decide to "stop" at 5 meters? $\endgroup$ – yoni0505 Jan 24 '18 at 10:56
  • $\begingroup$ You can stop applying force after 5 meters. The object won't stop, but it will cease to accelerate. $\endgroup$ – Allure Jan 24 '18 at 11:02
  • $\begingroup$ @yonio505 Like Allure said, acceleration is the key. Force is very closely related to acceleration (F=ma), so when an object stops accelerating you know that force is no longer applied to it. In free space the object would continue to move at the same speed indefinitely, but it would not accelerate. Thus, force is no longer applied to it. $\endgroup$ – Inertial Ignorance Jan 25 '18 at 8:35

protected by Qmechanic Jan 24 '18 at 9:12

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