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Discussion I am having:

If there are two slides that are at the same height. One slide is 100m long and the other slide is 200m long. The endpoint and start point are the same displacement. In a perfect environment with no friction, if two people went down at the same time with the same force, would they reach the bottom at the same time?

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closed as off-topic by ACuriousMind, Norbert Schuch, Kyle Kanos, Gert, JamalS Feb 14 '16 at 17:50

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  • $\begingroup$ What do you mean by "the same force"? Is there some force pushing them down the slide? Because if the only forces acting are gravity and the normal force exerted by the slide on the person, then the net force on the people sliding down the two slides must be different (it depends on the geometry, which must be different in the two cases). Please be more specific about details, and explain what aspect of this problem is causing you difficulty (what physical concept are you having trouble with here?). $\endgroup$ – march Feb 11 '16 at 14:58
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    $\begingroup$ Please clarify if the endpoint-startpoint displacement refers to vertical displacement (height) or otherwise $\endgroup$ – gregsan Feb 11 '16 at 15:13
  • $\begingroup$ There is no reason why the slide should be just twice as long. Lets create an absurd example where the length is so long, you have a very very slight incline, say a full kilometer. There is no friction in either case but in one case you're moving very slowly like you were sliding down ice and in the other, you're falling rapidly. Clearly length of the slide is significant. $\endgroup$ – Neil Feb 11 '16 at 16:11
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So this depends very strongly on the shape of the slide.

The easiest way to see this is to push it to its extreme: suppose one slide is purely vertical and has a length of 100 meters (i.e. $H = L$, then in the absence of friction getting to the bottom requires a free-fall time, which is gotten by solving $H - \frac12 g t_1^2 = 0$ to get a time $t_1 = \sqrt{\frac{2H}{g}}.$

The straight slide which is twice as long makes a $\theta = 30^\circ$ angle with the ground as it is half of an equilateral triangle. Rotating into its reference frame, we find out that gravity now points along the direction of motion with an acceleration of only $g\sin\theta$, which in this case is $\frac{g}{2}$. So now we have to solve $\frac12 \left(\frac g2\right) t_2^2 = L = 2H$

yielding $t_2 = 2 ~ t_1$ in the end.

There is a related misapprehension that students have: Now furthermore consider a bent slide, which travels straight down for this distance $H$, then conserving energy it bends the rider sharply to go forward, which they also go for a distance $\frac{\sqrt 3}2 H$ before leaving the ride. They get to the same place that the straight slide above got -- but does that take the same time?

We know that the "free fall" part takes time $\sqrt{2H/g}$ and using $v = -g t$ we can calculate that their speed at the bottom is $|v| = \sqrt{2Hg}.$ Preserving kinetic energy will come if the sharp turn preserves their speed, so now we just need to calculate the time to go that extra distance. And this time is $\ell/v,$ which in this case is $$t_3 = t_1 + \frac\ell v = \sqrt{\frac{2H}{g}} + \sqrt{\frac{\frac34 H^2}{2 H g}} = \sqrt{\frac{2H}{g}}~\left(1 + \sqrt{\frac 3 {16}}\right).$$ This number $1 + \sqrt{3}/4$ is only 1.433, much smaller than the $2.0$ that we saw for $t_2$ above! Same endpoint, different shape, got there much faster.

Now take the reverse path -- go horizontal from the starting point for a distance $H\sqrt{3}/2$, then go straight down. How much time does that take? $\infty.$ Because you never have gravity "start pushing you forward"!

One way to see this directly is to see that in the presence of a potential energy field $m g h$ like gravity has, your velocity is always going to be generated by energy conservation, $|v| = \sqrt{2 g (y_0 - y)}.$ You can think of this abstractly like a topographical map of sorts, with lower heights marked with higher velocities. A path which spends more of its time in the higher-velocity part of the map of course travels faster from point A to point B in the same time, until it gets overburdened by the distance that it has to travel. (The other $t \to \infty$ path is "the person falls down the slide an indefinite length $Q$, then sharply turns horizontal so they travel the $H\sqrt{3}/2$ at super-fast-speeds, then they sharply turn up so they travel the $Q-H$ distance to get to the same endpoint. As $Q$ gets larger this path becomes more about the "falling" and "rising" than the effectively 0 time spent travelling between.)

There is a smooth curve in between which is the fastest way to get from point A to point B by gravity alone, and it is a cycloid curve. Deriving it usually is done in a "calculus of variations" course.

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That's a good question, and the answer is no, it depends on the shape on the slide. Then the question you can ask is which shape gives the shortest travelling time on the slide ?

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Ignoring friction, final speed at the bottom will be same for both because they loose same potential energy (assuming same mass). Therefore, acceleration down the flatter path will be less, and it will take longer time to attain final speed.

In other words, steeper the path, faster is the downward acceleration. Longer path, (irrespective of shape of the path) translates into over all flatter path, which means less acceleration and longer time.

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  • $\begingroup$ Don't know who did a drive-by downvote, but they're wrong. Acceleration and impulse are important here. $\endgroup$ – Bill N Feb 13 '16 at 18:13
  • $\begingroup$ Some people just like to down vote without going through the effort to understand the answer. I usually do not down vote unless I see a clear flaw, Otherwise, if the explanation requires too much effort to understand, I do not vote at all. $\endgroup$ – kpv Feb 13 '16 at 20:07
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@ BowlOfRed. The question states would both objects come down at the same time with ideal environment, that is, not friction, same mass, and g=9.8m/s2. So think about it this way, here the only work done is going to be by gravity. W=Fdcostheta, the force would be gravity, therefore W=mgh. Since gravity is a conservative force, it does not matter what path the object is taking in the slide, because the work done by gravity depends only in the initial and final heights, not the path taken. We can go and check Fn, and yes we will have to take into consideration the vertical and horizontal angles as someone above responded. But again, he is just asking if they both get down at the same time with no friction, and since work done is only by gravity, the answer is yes.

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  • $\begingroup$ Amount of work done has nothing to do with the time it takes to do the work. The specific path is absolutely important in determining the total time of travel. $\endgroup$ – Bill N Feb 13 '16 at 4:39
  • $\begingroup$ The work done is indeed the answer to this problem. Work done by conservative force is independent on the path taken, it depends only on initial and final position. Assuming there is no friction, and the only force is gravity, then the speed at the bottom is going to be the same, v=(2gh)^1/2, and the time it takes from top to bottom t=(2h/g)^1/2, given they both start from rest, from same h, no matter what length the slide is. So, yes, they will get down at the same time with the same speed $\endgroup$ – Shizell Cordon Feb 13 '16 at 17:22
  • $\begingroup$ Consider two paths between $(0\,m, 1\,m)$ and $(3\,m, 0\,m)$. Both paths end horizontal at the end point. Two identical particles slide frictionlessly, one on each path, starting from rest. Path 1 is tilted at 1 degree below the horizontal until it is at $x=2.7 m$ at which point it drops precipitously with smooth, short transitions to its horizontal approach at $x= 2.95\,m$. It takes 5.62 s to reach the first transition. Path 2 is a straight line slope to $x= 2.95\,m, y=0$. It takes 1.41 s, is travelling 4.43 m/s, 0.05 away from end. Particle on path 1 isn't even half way along its path. $\endgroup$ – Bill N Feb 13 '16 at 18:44
  • $\begingroup$ Work is independent of time! AND, btw, it was not I who down-voted you. I would have told you. I hate drive-by downvotes. It doesn't help anyone. $\endgroup$ – Bill N Feb 13 '16 at 18:44
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Friction less environment - do you intend to say vacuum. If vacuum and if both went down the two slides of equal length, they reach the bottom at same time irrespective of their mass as the only accelaration is due to GRAVITY. Considering they both are at different lengths, the time taken will be greater if greater the height or length. Reason being the vacuum eliminated the mass but it did not change the distance to be traversed.

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    $\begingroup$ Gravity isn't the only force, there's a normal reaction due to the slide ; even if there is no friction. $\endgroup$ – Dimitri Feb 11 '16 at 15:25
  • $\begingroup$ Harsha, what if the only difference in two paths is one path drops below the first one for a reasonable distance then rises back up to match the first path and they both finish at the same point? The particle on the 2nd path will arrive sooner! It's an interesting problem, but consider the plot of $v_x$ versus time for each. The areas under the curves must be the same. $\endgroup$ – Bill N Feb 13 '16 at 18:58
  • $\begingroup$ Bill, that is an interesting question indeed. And your case is something I did not consider. My intrepretation is based on the paths being straight without any angular shifts. $\endgroup$ – Srih Feb 14 '16 at 10:30
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This is actually really simple. Yes! Both people will get to the end at the same time. In a perfect environment (say mass is the same, no friction force, and earth gravity), time is not a factor, and therefore both will come at the same time. The only forces acting on the people are $F_g$ and $F_n$, where $F_g = mg$ and $F_n$ really doesn't play a role so we neglect it. Now $F_g$ is a conservative force, that is, it does not take into account the path taken, just displacement. (Displacement is $d_f - d_i$.) No mention of time so far right? Since both objects have same mass and same gravity ($9.8\ \mathrm{m/s^2}$), and the displacement is the same, they both get down at the same time.

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  • $\begingroup$ Hi SharinH and welcome to Physics Stackexchange. I edited your post to use the mathematical markup we have here. If you want more gory details for what you can do with this, look here. $\endgroup$ – user10851 Feb 12 '16 at 3:34
  • $\begingroup$ Why do you think $F_n$ doesn't play a role? $\endgroup$ – BowlOfRed Feb 12 '16 at 21:29
  • $\begingroup$ The net force is providing the impulse. The net force depends on the path (that's what @BowlOfRed is pointing out.. We know that the total impulse must be the same. If the net force is different because a path is different, then the time interval of the impulse integral must be different. Total work can be the same because the work is independent of time. $\endgroup$ – Bill N Feb 13 '16 at 18:50

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