What is the Nature of Force in Newton's Second Law?

AI Thread Summary
Newton's Second Law, expressed as F=MA, indicates that a force applied to an object results in acceleration proportional to its mass. The discussion centers on whether the acceleration is ongoing or a one-time effect; it clarifies that as long as a net force is applied, the object will continue to accelerate. If the force is only applied momentarily, the object will stop accelerating once the force is removed. The participants emphasize that force is not persistent but rather instantaneous, ceasing when the push is no longer applied. Understanding this concept is crucial for grasping the dynamics of motion in physics.
godingly
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Homework Statement


Thanks for looking. My understanding of Newton's Second Law, F=MA, is: If I give some mass a force, it will accelerate at some rate.
Imagine an infinite, frictionless surface. If I give a box of 1kg a push of 6 Newtons, it will accelerate at 6 m/sec2. So, at t=0, v=0. at t=1, v=6. I have two questions:
1) What is v, at t=0.5? is it 3? is it 6?
2) what is v, at t=2? does it remain 6, or continue accelerating to 12? and why?
What I'm trying to ask is, is the A in F=MA is an ongoing acceleration, that keeps accelerating the box as long as there is no friction, or is it a single acceleration, giving an addition of 6 m/sec2 once, and that's it?



Homework Equations


F=MA


The Attempt at a Solution


Asked my sister and someone who graduated in physics, read the wikipedia article, searched Google, read relevant chapters in Conceptual Physics, and watched Prof. Walter Lewin (Best Accent!). No one addresses my questions. Thank you very much!
 
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godingly said:
What I'm trying to ask is, is the A in F=MA is an ongoing acceleration, that keeps accelerating the box as long as there is no friction, or is it a single acceleration, giving an addition of 6 m/sec2 once, and that's it?
As long as there is a net force on the box it will continue to accelerate. From the definition of acceleration (a = Δv/Δt) you can figure out the speed at any time. (Since the acceleration is uniform.)
 
Thank you, Doc Al. I will restate my question: Assuming I give the box a push, and stay in my place (or drift to the opposite side, according to the third law) :
Is the force I gave to the box is ever-present, or is it instantaneous? if at t=0, f=6, what is f at t=0.5, or t=1? if it's zero, what opposing force eliminated the push? or does force "evaporates" the minute you use it?
 
The force on a system vanishes immediately when it is removed from the system.
 
godingly said:
Thank you, Doc Al. I will restate my question: Assuming I give the box a push, and stay in my place (or drift to the opposite side, according to the third law) :
Is the force I gave to the box is ever-present, or is it instantaneous?
The force exists for as long as you keep pushing! Newton's 3rd law says that as long as you keep pushing the box, the box will keep pushing back on you. (So what?)
if at t=0, f=6, what is f at t=0.5, or t=1? if it's zero, what opposing force eliminated the push? or does force "evaporates" the minute you use it?
As soon as you stop pushing, the acceleration becomes zero. If you just give the box a single shove, thus exerting the force for some time (and not forever), the box stops accelerating as soon as you stop exerting the force.

Perhaps you can restate your question, as I'm not sure I'm capturing what you have in mind.
 
I thought that unless an opposing force has acted on the box, the net force on the box will continue to be 6 Newtons, and therefore the box will continue to accelerate. But, the force is instantaneous, and "evaporates" immediately. no textbook tells you that about the nature of force. Got it.
Thanks you Doc Al, and Thank you physicsisgrea!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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