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First Question: Force applied for a span of time.

Let's say a force $F$ is applied on an object of mass $M$. So, the acceleration of that object in the direction of $F$, would be, $a = F/M$.

Now, if we apply the force for 5 seconds, what will be the acceleration on the object? Will we take the net force exerted on the object to be $5F$ and thus the acceleration would be $5a$?


Second Question: Objects falling after thrown

Let's say I have a ball on my hands. Right now, the only acceleration being applied on the force is the gravitational acceleration $g$, which is downwards (towards to ground). Now, if I apply force $F$ on the ball directly upwards, creating acceleration $a$ in the ball, the net acceleration upwards will be $a-g$.

If I were to be in a vacuum without any air resistance, the ball would keep moving upwards with acceleration $a-g$. But due to the air resistance (the ball hitting the air particles while trying to go upwards), the upwards acceleration $a$ decreases over time and at some point $a=g$ and $g>a$ afterwards. Then, the net acceleration would be downwards thus the ball will start falling to my hands.

Is this the right thought process or am I getting something majorly wrong?

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    $\begingroup$ How do you apply a force to the ball after it has left your hand? $\endgroup$ – PM 2Ring Jun 6 at 12:54
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    $\begingroup$ One question at a time, please. $\endgroup$ – Bill N Jun 6 at 17:50
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You are getting three things majorly wrong!

If you apply a constant force F to a mass M for one second, the mass will accelerate at a constant rate of F/M for one second. If you apply the force for five seconds, the mass will accelerate for five seconds. So increasing the duration of the applied force simply increases the duration of the acceleration by the same amount.

What you might have had in mind is that if you apply a force of F for five seconds you achieve the same overall change in the velocity of the mass as you would had you applied a force of 5F for one second. However, the two situations are not directly comparable.

The next majorly wrong thing is that when you hold a ball still in your hand, gravity is not the only force acting on it. Gravity is pulling the ball down, so your hand must apply an equal and opposite force to resist gravity, with the result that the net force on the ball is zero.

If you then apply an additional force to accelerate the ball up in the air, the acceleration due to that force stops as soon as the ball loses contact with your hand. From that point onwards, gravity acts to accelerate the ball downwards. Gravity reduces the ball's upward speed until the ball stops altogether momentarily, then gravity increases the ball's downward speed. The air resistance is a red herring.

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  • $\begingroup$ Ah yeah you are absolutely correct. I completely messed up my kinetics concept while studying other topics. You pretty much solved my doubts. Also in this part " gravity is not the only force acting on it. " , I knew about this so I said that " the only acceleration acting on the ball is ". I think you might've misread. Anyway, I also had the following stuff majorly incorrect so thank you for pointing it out. If anyone else is also having issues with the same thing as I , the thing you have to understand that Acceleration is only when there is Force acting on the object. $\endgroup$ – user295079 Jun 6 at 14:00
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Will we take the net force exerted on the object to be $5F$ and thus the acceleration would be $5a$?

No this is incorrect. If the net external force applied to the object is $F$ and is constant, then irrespective of how long the force has been applied, the acceleration of the object would be $a = \frac{F}{m}$. I think you are getting confused with the concept of velocity and acceleration. Because the force applied is constant, the acceleration of the object would then be constant, which would lead to a constant increase in the magnitude of the object's velocity. This is governed by the particle under constant acceleration model ; If it had no initial speed, then $$v_f = v_i + a\cdot t \Rightarrow v_f = a\cdot (5\rm s)$$ where $\rm s$ refers to the seconds, and not distance.

Right now, the only acceleration being applied on the force is the gravitational acceleration $g$, which is downwards (towards to ground)

"Acceleration being applied on the force" is meaningless. If the ball is on your hands, it is at rest, and is not accelerating because its velocity is constant (zero). This is because apart from the gravitational force, your palm also exerts an upward contact force which causes it to experience no net force, and hence does not accelerate in any direction.

Now, if I apply force $F$ on the ball directly upwards, creating acceleration $a$ in the ball, the net acceleration upwards will be $a−g$

When you throw a ball, you do not impart acceleration into it. In fact, the moment it leaves your palm, the only force it experiences is the gravitational force, that points downward. Because you accelerated the ball while it was on your palm, it has gained some initial velocity. As the only force exerted on it causes it to accelerate downward, that upward initial velocity continuously decreases until it reaches zero (at its maximum height) $$v_f = v_i - gt$$ The force doesn't disappear then, it still acts on the ball, and the downward acceleration causes it to increase its velocity in the downward direction.

Hope this helps.

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Let's say I have a ball on my hands. Right now, the only acceleration being applied on the force is the gravitational acceleration g, which is downwards (towards to ground).

This is a major conceptual misunderstanding. Acceleration is not something which is applied, or acts, or causes anything. Acceleration is a part of the description of the kinetic behavior of an object.

  1. If an object changes position within a reference frame, the rate of change of position is called velocity, relative to that frame.
  2. If an object's velocity changes within a reference frame, the rate of change of velocity is called acceleration.

Acceleration simply tells us what is happening to an object. It doesn't cause an object to move, nor does it cause a change in velocity. It tells us whether or not the velocity is changing.

Newton's second law describes the relationship between the (observed, or a possible) acceleration, the mass of an object, and the forces applied to an object.

Finally, "gravitational acceleration" is, at best, an extremely misleading term because it merely describes a situation in which gravity is the only force acting on some object. If there are other forces acting, there is no gravitational acceleration. The quantity $m\vec{g}$ is a force, and $\vec{g}$ is a gravitational field usually due to some planet, not an acceleration. It is, in Newtonian physics, $$|\vec{g}|=\frac{GM_p}{r^2},$$ where, $G$ is the universal gravitational constant, $M_p$ is the mass of the ideally spherical planet, and $r$ is the distance from the ideal center of the planetary mass to the point at which $g$ is being calculated.

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