eli5 Force and work equaling zero

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So if force =mass * acceleration, then an object moving at a constant velocity has zero acceleration, and thus force = zero.
So is f=ma actually mean net force or some other definition?

And I’ve been told work is force * distance, so sn object moving at a constant velocity would also have zero work? Which doesn’t sound correct so I’m confused on what exactly work defines

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Anonymous 0 Comments

Hi /u/AwesomeGuyDj!

>So is f=ma actually mean net force or some other definition?

Yes, the acceleration of an object is given by the net force on said object. Since force is a vector, and vectors can be simply added, the net force is the sum of all the forces on the body. In mathematical notation, this is given by the following equation: **F_net** = Σ **f_i** = m · **a**; where **f_i** are the forces on the body, and **bold** objects are vectors.

>And I’ve been told work is force * distance, so sn object moving at a constant velocity would also have zero work? Which doesn’t sound correct so I’m confused on what exactly work defines

Assuming the net force is constant and we are in the rest frame of the object at t=t0, the work W done on some object is indeed given by W = F · s, or the distance s over which the force is applied multiplied with the absolute value of the force. F

While this might seem counterintuitive, consider a universe that is empty except for some object with mass m. When the object is moving at a constant velocity relative to some frame of reference I, no force is required to keep it moving at that constant speed, since an object in motion will stay in motion, unless a force is acting on it. As no force is required to keep it moving, no work is done on the object: W = F · s = 0N · s = 0J (Since F=0N).

The origin of your confusion might be, that this thought experiment does not coincide with our experience: in every-day life, an object in motion does *not* stay in motion, right? But here, we have to remember that the force of friction *is* acting on the moving object, thereby slowing it down.

Does that help?

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