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I know $F=ma$, but it's never really made sense to me. Here's an example:

Hypothetical scenario- I push a bowling ball ($m = 4\,\rm kg$) sitting on the floor with $5\,\rm N$ of force. (Assume no friction or air resistance.) Note that I do not continue to push the bowling ball. I only push it once, with the initial force of $5\,\rm N$. The bowling ball supposedly moves, that is, it now has a velocity. Let us assume that the ball moves with a constant velocity immediately (equal and opposite reaction because the $5\,\rm N$ of force was applied instantaneously (right?)). But if the ball is moving with a constant velocity, then its acceleration would be zero. Therefore, the force applied should've been zero.

Simple question: If force causes velocity, then why is force not related to velocity?

Reason for question: This has to do with applying a force to an object to make it go up. It's a King of the Hammer type configuration, where you use a hammer to smash one end of the lever, applying a force to a metal block to raise it into the air. How much force is required to raise a $500\,\rm g$ object $0.2\,\rm m$ into the air if it rests on a $0.1\,\rm m$ lever, the fulcrum being $0.04\,\rm m$ from the object?

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  • $\begingroup$ Almost nothing in your hypothetical scenario makes sense. “Do not continue to push the bowling ball. I only push it once” is bizarre. It doesn’t make sense to count the application of a 5 N force. You can apply a 5 N force for 1 s or over 1 m, but once makes no sense. The rest is similarly confusing. $\endgroup$ – Dale Oct 31 '18 at 2:39
  • $\begingroup$ You did make enough sense for me to understand what was meant, but as Dale seems to be pointing out, you should consider trying to be more precise with your language when using this forum, OP. $\endgroup$ – PiKindOfGuy Oct 31 '18 at 2:43
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The answer to your question is that in $F = ma$, $F$ and $a$ are taken at the same time $t$. Your example fails to refute Newton's second law because you're considering the acceleration after you pushed the ball. Force is related to velocity through acceleration: a force causes an acceleration which causes a change in velocity (in this example, from a velocity of zero to something else).

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  • $\begingroup$ So if I apply a 5 N force on a 4 kg object, it would have an instantaneous acceleration of 1.25 $m/s^2$ which falls to zero post-force-application, and therefore maintain a velocity of 1.25 $m/s$ for an indefinite time t? $\endgroup$ – Christopher Marley Oct 31 '18 at 2:38
  • $\begingroup$ @ChristopherMarley Correct! $\endgroup$ – PiKindOfGuy Oct 31 '18 at 2:39
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    $\begingroup$ @ChristopherMarley If that force is applied for $1\ \rm s$ then yes. $\endgroup$ – Aaron Stevens Oct 31 '18 at 2:42
  • $\begingroup$ What Aaron said. Yes, if it were applied for two seconds, you'd end up with a velocity of 2 * 1.25 m/s = 2.5 m/s. $\endgroup$ – PiKindOfGuy Oct 31 '18 at 2:45
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If you want to relate force to velocity instead of acceleration, it's not very hard to show how and why velocity is affected.

We have $F = m a$, but we also know acceleration is just a change in velocity over time.

That means we can say $a = \frac {dv}{dt}$ (if you're not familiar with deravitaves, it's basically saying at any point in time, the acceleration is just the slope of the velocity-time graph).

If you substitute that in, you get: $$F = m \frac {dv}{dt}$$

If we simplify it a little bit by saying $dv \approx \Delta v$ and $dt \approx \Delta t$, we can rearrange to get: $$F \Delta t = m \Delta v$$

In other words, the change in velocity of the object depends on both the amount of force, and the time it takes to apply the force. Another way to think about it is that force is the change in momentum over time.

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