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I know this is a very simple question, but I am just learning physics. I am seeing the basics of how a block on a horizontal frictionless surface gets accelerated by a force F in any direction. However, I ask myself if F=ma is only used on the object the force is being applied to, not the source of the force (which could be a finger, another block, or anything that pushes)

Could one use F=ma in order to calculate the force an object of mass M would produce on its own if it already had an acceleration. Thank you for you helpful responses in advance

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4 Answers 4

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It's great that you're asking conceptual questions rather than just plugging in to formulas.

Does it mean that an object with mass M will have an acceleration a if an external force is applied to it?

Yes.

However, I ask myself if F=ma is only used on the object the force is being applied to, not the source of the force (which could be a finger, another block, or anything that pushes)

Right, it's only used that way, not for the source.

Could one use F=ma in order to calculate the force an object of mass M would produce on its own if it already had an acceleration.

No.

(1) It doesn't make sense to talk about an object already having an acceleration. When we talk about what an object already has, we're describing its present state. Its state is the information that, if the object was left alone, we would later be able to reconstruct. It retains information about its position (x) and its velocity (v). It doesn't retain information about accelerations. For example, if I see a hockey puck sliding across the ice, I can see its current x and v, and I can also reconstruct its past x and v, going back in time for as long as it was being left alone. I can't reconstruct anything about its past accelerations.

(2) A force is always an interaction between two objects, so it doesn't make sense to talk about an object having a force on its own.

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  • $\begingroup$ Thank you very very much!! $\endgroup$
    – AB2
    Commented Jan 22, 2021 at 13:38
  • $\begingroup$ if the object had this acceleration and the applied force suddenly dissapears, then the block would continue with a constant velocity that it managed o achieve while accelerated right? $\endgroup$
    – AB2
    Commented Jan 22, 2021 at 14:07
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If the object already has an acceleration, then yes there is (total) force acting on it given by $$F=ma$$ It does not matter where the force is coming from: if the total force acting, of whatever origin, is $F$ then the object's acceleration is going to be $ma$. If you have multiple forces acting at the same time (e.g. you are pushing together with another person) then you have to sum such forces.

However, it is not clear what you mean by "a force produced on its own". Masses don't "produce" force on their own. If you have a block of some material moving with acceleration $a$ there must be a force pushing it, that's all you know.

If you are referring to a car, or a motor, or something able to propel objects, then yes, the force $F$ may be due to some process happening "inside" the object. But what is actually pushing the object is usually a reaction force due to Newton's third law (action / reaction): for example, if you are walking, you are pushing on the ground and the ground is pushing you back. So the force you have to consider is the force acting on the object, not the force the object is exerting.

Some examples (all objects in these examples have mass $m$):

  • an object is falling down with acceleration $a$. Then gravity is exerting a force on the object given by $F=ma$

  • you push an object with force $F$ on a surface with friction and notice that the object moves at constant speed ($a=0$). This means that friction is counter-exerting a force $F-f$ exactly the same as yours but in opposite direction, so that the total force is $F_{tot}=F-F_f=ma=0$

  • a car is moving with acceleration $a$. This means that the car is exerting some force on the ground through the wheels such that the ground is responding by answering with a force $F$.

As you keep studying, you will learn how to describe all of such processes, but the summary of this answer is:

in $F=ma$ the force $F$ in this equation refers to the sum of all the forces acting on the object

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  • $\begingroup$ Thank you a lot! $\endgroup$
    – AB2
    Commented Jan 22, 2021 at 13:39
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However, I ask myself if F=ma is only used on the object the force is being applied to, not the source of the force (which could be a finger, another block, or anything that pushes)

$F_{net}=ma$ is Newton's second law. Newton's third law says that the object the force $F$ is applied to exerts an equal and opposite force $F$ on the object applying the force (finger, another block, or anything that pushes). The effect on the object applying the force will depend on the net force it experiences, per Newton's second law applied to it.

Could one use F=ma in order to calculate the force an object of mass M would produce on its own if it already had an acceleration.

"...had an acceleration" implies that the object of mass M is no longer accelerating, that is, it is no longer being subjected to a net force, i.e., $F_{net}=0$. It is therefore continuing at the velocity it had when the force was removed and therefore neither producing or experiencing a force (assuming no friction or air resistance). However such an object could produce a force on another object if collided with another object.

Hope this helps.

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The answers here are already great, but here's a small suggestion I can add: when learning Newton's laws, I think it's easier to start off by thinking of Newton's second law as $$ a = \frac{F}{m} $$ This is mathematically the same as $F = ma$, of course, but I've noticed that students tend to think that an equation $X = Y$ means that $Y$ causes $X$ -- likely because they are use to plugging in values on the r.h.s. to get the value on the l.h.s. I think it's this that's behind your question about whether $F=ma$ means a mass with acceleration can generate a force. This isn't the right causal order, though; it's the force that causes the acceleration. It's crucial to remember this and be familiar with it when solving problems, and thinking of the law as $a = F/m$ might help with that.

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  • $\begingroup$ Thank you a lot! It makes much more sense now. $\endgroup$
    – AB2
    Commented Jan 22, 2021 at 15:24

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