0
$\begingroup$

I am somewhat confused by f =ma This means a = f/m

But there is no time associated with the force to say how long the force is applied to give us this acceleration or how long the body will accelerate if no friction is opposing the force.

So if we apply the force f for an infinitely small amount of time or follow the object as in shove it for a bit longer the acceleration is the same? (By follow I mean the force stays in contact with the object for longer)

But the energy expelled or work done is not the same. Since force times distance force moves is different when does it stop accelerating, it's like we have derived the instantaneous acceleration.

$\endgroup$
3
  • 2
    $\begingroup$ What exactly is your question? $\endgroup$
    – M. Enns
    Commented May 16, 2021 at 21:48
  • $\begingroup$ do you know differential equations? $\endgroup$
    – user65081
    Commented May 16, 2021 at 22:39
  • $\begingroup$ The work is not the same, but in the case where you follow, the final kinetic energy is larger than in the short-duration case. $\endgroup$
    – garyp
    Commented May 17, 2021 at 1:14

3 Answers 3

1
$\begingroup$

Not exactly sure what your question is, but I believe it is related to confusion about acceleration.

Acceleration is itself an "instantaneous" concept. Acceleration is the rate of change of velocity and velocity is the rate of change of displacement. Time comes into play when you solve for the velocity and displacement given the acceleration. Acceleration is force- possibly time dependent- divided by the constant mass.

Acceleration $\vec a ={{d \vec v} \over {dt}} = {{d^2 \vec r} \over {dt^2}}$ where $\vec a$ is acceleration and $\vec r$ is displacement. $\vec a = {\vec F \over m}$ where $\vec F$ is force.

You have to integrate the relationship for $\vec a$ to determine $\vec v$ and then integrate that relationship for $\vec v$ to determine $\vec r$. Consider a constant force, $F$, acting in one direction (call it x) on a particle initially at rest over a time $t$. The acceleration is constant during that time, $a = F/m$, and by solving the relationships in the first paragraph, the total displacement is ${1 \over 2} at^2$.

Work is $\int_{r_1}^{r_2}\vec F \cdot d \vec r = \Delta KE$ where $\Delta KE$ is the change in kinetic energy. With no force the kinetic energy, hence the velocity, is constant.

$\endgroup$
0
$\begingroup$

Acceleration is indeed instantaneous (a does not depend on its history). I think what confuses you is that in the real world this rarely happens. This is because in the real world, F usually does not change instantaneously. For example, when you push a box to let it accelerate, your arm will need to move forward. Since position cannot change instantaneously, it takes time for your arm to gradually increase the force on the box. However, F can indeed change instantaneously in certain scenarios, in which case the acceleration will change instantaneously, too. For example, electromagnetic force can change instantaneously by controlling the electrical current. Therefore when you drive a Tesla, you feel the acceleration very much "instantaneous" (of course, there is still some small lag because you need to move the position of your feet to push the pedal first).

$\endgroup$
0
$\begingroup$

These simply are truths of how force and acceleration work. They may be odd, but they are right.

Force and acceleration do not depend on how long these things are applied. 9.8m/s^2 is 9.8m/s^2 whether it is applied for 1 ns or a century.

Now, if you are not familiar with calculus, this can seem a bit wonky. The idea of there being an acceleration for an infinitesimal period of time is kind of awkward. Newton and Leibniz invented calculus to deal with these. In fact, when they were first proposed, the name was "the calculus of infinitesimals," where "calculus" refers to a method of calculation (indeed: same root as calculation and calculator). Until they came along, we didn't have a rigorous way of dealing with infintessimals. Zeno's paradox about how an arrow cannot hit its target because it first has to get half way there, and half way to the half way point and so forth, was troublesome.

So until you have calculus under your belt, it will seem a little strange. Force will feel like something instantaneous, and acceleration will feel like something that's done over a period. But if/when you learn calculus, you will learn that we treat both of these things as instantanious.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.