A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is?
Answer to this question is only g. But why does a does not influence the stone.
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$\begingroup$ It depends on what reference frame you are talking about. If in the lab frame, the acceleration of the stone is indeed $g$. When the stone is initially released, it will be in free fall and not in contact with the elevator. Therefore, the only force acting on the stone is it’s weight. $\endgroup$– Ashmit DuttaCommented Jul 1, 2022 at 4:05
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$\begingroup$ @AshmitDutta can acceleration become zero at an instant. or does it decrease slowly. $\endgroup$– Dron BhattacharyaCommented Jul 1, 2022 at 4:39
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$\begingroup$ Should ground frame be considered whenever reference frame is not mentioned $\endgroup$– Dron BhattacharyaCommented Jul 1, 2022 at 4:40
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1$\begingroup$ "can acceleration become zero at an instant" - well, you can't release a stone instantaneously, so there will be some function that describes "current" acceleration supplied to the stone by your hand, that drops off rapidly as you release your grip, but as soon as the contact is broken, it'll be zero (apart from g). There is no "lag", the object doesn't carry the acceleration with it - if that's what you're asking - as it is imparted on it by external forces. It's the velocity that'll continue on from the value it had at release time, and then change slowly because of g. $\endgroup$– Filip MilovanovićCommented Jul 1, 2022 at 5:18
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1$\begingroup$ @FilipMilovanović Now it is clear to me. When the stone is not yet released, acceleration of the elevator (a) is acting on it. When the stone is released, a does not influence the stone anymore. Now acceleration due to gravity is only acting on it in the downward direction. So, the velocity gained in upward direction before released is now decreasing slowly. Hence, acceleration does not decrease slowly. This is what I have understood from your comment. $\endgroup$– Dron BhattacharyaCommented Jul 1, 2022 at 5:27
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2 Answers
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It's simply Newton's second law. What are the forces acting on the stone after its release? Well, it's completely out of contact with the elevator, so there's only the force of gravity, meaning that $m\mathbf{a}_\text{stone} =\mathbf{F} = m\mathbf{g}$, i.e. $\mathbf{a}_\text{stone} = \mathbf{g}$.
Of course, I'm assuming that we're talking about the acceleration of the stone in the reference frame of the earth. The relative acceleration between the stone and the elevator would be $\mathbf{a}_{\text{stone/elevator}}=\mathbf{g}-\mathbf{a}$.
answered Jul 1, 2022 at 4:17
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The elevator is accelerating because there is some force acting on it (presumably the cable). While you are holding onto the stone, that force is transmitted through to the stone via the floor, your body and your arm, so the stone accelerates also at the same rate as the elevator.
As soon as you let go of the stone, the force from the cable no longer acts on the stone. The lift carries on accelerating (presumably slightly more now, since the cables don't have to accelerate the stone as well), but the stone is free to fall under gravity only.
Hence, in the reference frame of the outside world, the stone only falls under gravity ($g$).
In the reference frame of the elevator cabin, the magnitude of the acceleration will indeed be $a + g$.
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$\begingroup$ Is your last line missing a minus sign? $\endgroup$– WillOCommented Jul 1, 2022 at 12:28
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$\begingroup$ Well, technically (although I am not quite sure what you mean). The cabin is accelerating up, so the stone will appear to be accelerating down from the reference frame of the cabin. This is in addition to $g$ so the magnitude would be $g+a$. However, since the direction is towards the floor, you could say $-(g+a)$. I will add the word "magnitude". Here, I am assuming the convention of $g$ being positive. $\endgroup$– rghomeCommented Jul 1, 2022 at 12:36
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$\begingroup$ Actually, I was mistaken. (I was thinking the g and a went in opposite directions, which is clearly wrong.) My apologies. $\endgroup$– WillOCommented Jul 1, 2022 at 12:40