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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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?
That analysis is mostly correct although perhaps it makes it easier to think of it not as “work” but “work done”. When thinking about physics problems, the important thing is NOT to switch contexts.
What this means is that the object that is moving at a constant velocity is not doing work or has work being done on it. So FROM THE PERSPECTIVE OF THE OBJECT, there is no gain or loss of energy (for an ELI5 the energy is the work content of the object)
In:
> **F** = m **a**
The “F” is the resultant force, or the net force, or the vector sum of all forces.
Work done on an object as a result of a specific force is the force * distance (or if we’re being more precise, the path integral of **F.ds**).
So these are different forces. The first is the net force, the second is a specific force.
Quoting Wikipedia, “work is the energy transferred to or from an object via the application of force along a displacement.”
So as you note, if an object is moving at constant velocity, there must be no net force acting on it – i.e. all forces must cancel each other out. So there is no overall work done on or by the object. This is the **work-energy principle**:
> the work done by all forces acting on a particle (the work of the resultant force) equals the change in the kinetic energy of the particle
Here we have no change in kinetic energy, so we have no net work done.
Where things get a bit complicated is that work doesn’t just transfer energy to an object from an object, it can also shift energy from one form to another form.
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If we take the example of lifting an object up a certain height *h* at constant velocity, there will be two forces acting on it. Some force that is lifting it up, *L*, and its weight pulling it down *W*. The work done on the object by the lifting force will be:
> w.d. = L * h
The work done by the weight will be:
> w.d. = W * h
(noting that W will be negative if we are taking up to be the positive direction). Then by F = ma, if it is at constant velocity we know *a* = 0, and so:
> F = L + W = ma = 0
> L = -W
and so we get that the work done on the object by the lifting force is equal and opposite to the work done by the weight; so there is no overall change in work on the object.
But the object *has gained energy* – it has gained gravitational potential energy. While the work done by the lifting force is transferring energy from whatever is lifting it to the object, the work done by the weight is shifting that energy into the form of gravitational potential energy (or alternatively, transferring that energy into the object-Earth system/gravitational field).
Potential energy is a bit of a fudge, in some ways. The maths works, but it can take a bit of thinking to get your head around it.
Any object will continue moving at a constant velocity in a straight line unless something does some work on it. If you want it to change direction, speed up or slow down, you’re going to have to apply some force to it. The bigger the distance you apply the force, the more work you do.
You are correct that the net force on an object is zero if the velocity is constant.
Let’s just look at the situation where our forces are directed towards the center of mass so they can result in an object starting to rotate around its own axis. We also assume a rigid body ie one that does not deform
Both acceleration and force are vector units so you can start by adding the forces together or do it at the end by adding acceleration together. Matematicaly it is equivalent because mass is a scalar and the samm for all.
An example of adding accelerations is if gravity is involved because the acceleration is independent of the mass but the force depends on the mass.
Work is the amount of energy that is transferred. But for it, you do need to look at the forces separately. Constant velocity if there is no friction air resistance etc requires zero force on an object.
But if you have friction, air resistance, etc it results in a force and will change the velocity, this is why a car will stop on flat ground if you turn off the engine. So the engine of a car provides a force that counteracts friction and air resistance and it does work because it is applied over a distance.
So in the practice of macroscopic objects on earth contact velocity will always require some work.
The answer to your first question is yes. f is the net force acting on an object.
The answer to your second question is also technically yes, however obviously work was done to get the object up to speed in the first place.
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Force is a definition. We define force as the change of momentum. Momentum is p=mv and the change so d/dt mv = ma as the change in velocity is acceleration and we define force to equal that ma. F:=ma.
An object moving with a constant velocity has no net force acting on it. It has momentum, if that momentum changes forces acted on the object.
Work is about doing stuff. If you want to pull a bucket from a vell then you need to do work on the bucket. If you shoot an arrow up with a constant velocity to get the bucket out, as the arrow pulls the bucket it will slow down, there is a change in momentum, there is a force, and the bucket moves up so work has been done. Work is the change in momentum over a distance. So you can apply a large force over a short distance or a small force over a longer distance the amount of work is the same.
So for the bucket example you have some wheel to turn to pull it up, pulling the bucket up you have to do work against gravity. If you pull the bucket up to height h. You need to do W=mgh amount of work. You can do it with a small wheel and then you’ll have to apply a lot of force. But with that force you will only need to turn the wheel a short distance. With a larger wheel as you apply more torque with it you apply a smaller force but over a longer distance. The result is the same Fs=mgh.
Yes an object just travelling with momentum p doesn’t do work, it does work when its momentum changes d/dt p. Which is what we defined a force to be. The latger distance that force is applied over the more work has been done.
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