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If I had a mass of $100\:\rm{kg}$ accelerating due to gravity, using $F=ma$:

$F = 100\:\rm{kg} \times 9.8\:\rm{m/s^2}$

$F = 980 \:\rm N$...

If I increased the mass to 200kg, the force would be 1960 N:

$F = 200\:\rm{kg} \times 9.8\:\rm{m/s^2}$

$F = 1960 \:\rm{N}$

Now, finally getting to my question: Does this increase in force (which is supposed to be a push/pull) mean that the object would fall faster when it weighs more?

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    $\begingroup$ I think You did nor really understand the meaning of "acceleration". $\endgroup$
    – Georg
    Commented May 17, 2011 at 11:34
  • $\begingroup$ I do understand that acceleration is linked to the change in velocity... however, to my mind, a stronger shove at someone or something would mean a greater velocity. So this was my confusion, whether I should follow the definitions or follow my (obviously incorrect) mind. $\endgroup$
    – edwardyu236
    Commented May 17, 2011 at 11:46
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    $\begingroup$ It is not "linked to", it is change of velocity per unit time!. So, because acceleration id not changed (always 9.8 m/s²) , why should velocity be changed when switching from 100 kg to 200 kg? $\endgroup$
    – Georg
    Commented May 17, 2011 at 11:49
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    $\begingroup$ @eswardyu236: "however, to my mind, a stronger shove at someone or something would mean a greater velocity." It doesn't. As Georg said, the change in velocity is "produced by" the acceleration, not the force. $\endgroup$
    – David Z
    Commented May 17, 2011 at 19:09
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/5973/2451 and physics.stackexchange.com/q/11321/2451 $\endgroup$
    – Qmechanic ♦
    Commented Oct 21, 2012 at 5:30

3 Answers 3

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No, the heavier object does not fall faster. Instead, they heavy and light object fall at the same acceleration (and hence the same speed if they are both simply dropped). This is an example of the equivalence principle.

The more massive object has more gravitational force on it, but it also has more inertia. Specifically, because the object is twice as massive, it has twice the inertial mass.

The force on it is doubled, so the acceleration stays the same.

If we look at

$$F = ma$$

we see that when $F$ and $m$ are both multiplied by 2, $a$ stays the same.

Check these questions for more:

Free falling of object with no air resistance

Why is heavier object more reluctant to get falling down?

Projectile motion without air resistance

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  • $\begingroup$ Thanks. If the object is on an inclined plane, the increased mass would increase the force without increasing the speed the object slides down the plane, right? $\endgroup$
    – edwardyu236
    Commented May 17, 2011 at 11:09
  • $\begingroup$ @edward No. If it's on a plane, increasing the mass increases the force, but again it still has more inertia. Things all slide down the same (frictionless) inclined plane with the same acceleration, just as they fall with the same acceleration. $\endgroup$
    – Mark Eichenlaub
    Commented May 17, 2011 at 11:16
  • $\begingroup$ however, on a plane the acceleration is smaller $\endgroup$
    – Mark Eichenlaub
    Commented May 17, 2011 at 11:19
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Agreed that the answer to the question is, "No." Acceleration remains constant.

One way to think of it is this: The first 980N is accelerating the first 100kg at 9.8m/(s^2). The second 980N is accelerating the second 100kg at 9.8m/(s^2). Both masses fall at the same velocity and acceleration (neglecting wind resistance, etc.). So it should be easy to see how you could join these to 100kg masses into one 200kg mass, with 1960N pulling it down, without changing the speed or acceleration of either mass.

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Lets assume there are no non-conservative forces like air drag and you are dropping the ball from rest. While the ball gains kinetic energy, it loses potential energy. This means that

$$ K_\text{in} + P_\text{in} = K_\text{f} + P_\text{f} $$ $$ 0 + mgh = \frac{1}{2} mv^2 + 0 $$

which means $v= \sqrt{2gh}$ and there is no mass term here. Hence the (final) velocity is not dependent on weight.

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