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I'm trying to understand the mathematics behind calculating the total speed and force of a $40 kg_f$ object hooked up to a parachute, falling to Earth.

From what I understand, the formula for this is $$m \frac{dv}{dt} = (mass \times gravity) -Force_{deceleration}= 0 $$

Here's where I'm confused: I understand that $m \frac{dv}{dt} = 0$ because once the parachute is deployed, the object reaches a constant speed at which it falls for the rest of the way down. However I also learned that $m \frac{dv}{dt} = F$, which is the total force of any moving mass. The object is falling to Earth and therefore it is moving, so how could the force, $F$, be equal to zero??

If you could help me get things straight, I'd really appreciate it.

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There are actually 2 forces acting on the object . 1) Weight = mass * gravity 2) Air resistance

At first the object is accelerating because the Weight > Air resistance, until the object reaches its terminal velocity where Air resistance = Weight, meaning there is no more acceleration, or in other words, a Net Force of 0.

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  • $\begingroup$ My question asks, how could the force be $0$ if the object is falling? Obviously if it's falling, some force is acting on it. $\endgroup$ – CodyBugstein Nov 8 '12 at 4:56
  • $\begingroup$ At the beginning there is a Force that is gravity, until it reaches its terminal velocity, take a friction-less surface, and say a if a ball is rolling on the surface, at constant speed then, then the ball will never stop or go faster, since there are no forces acting on it but it is moving still. $\endgroup$ – Dreamer78692 Nov 8 '12 at 5:01
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    $\begingroup$ Aha! That's a fascinating insight. So what you're saying is that is there is only a force in effect while the object is accelerating. Once it stops accelerating, that's when the application of the force has ended. I understand it now, but it wasn't clear initially in your answer. $\endgroup$ – CodyBugstein Nov 8 '12 at 5:09
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    $\begingroup$ @Imray I have edited the answer it is explained, sorry for the useless explanation at first. $\endgroup$ – Dreamer78692 Nov 8 '12 at 5:23

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