I obtain the following question in this link.

enter image description here

I don't understand why the velocity at the bottom for the four slides are the same. I thought the velocity depends on the angle of inclination between slide and the floor. In other words, my answer is

$$v_D > v_A > v_B > v_C$$

Can anyone explain to me why all speeds are the same?


closed as off-topic by knzhou, John Rennie, CuriousOne, sammy gerbil, heather Aug 4 '16 at 12:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ VtC voters: check the link, it is not homework. $\endgroup$ – user259412 Aug 4 '16 at 4:00
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    $\begingroup$ @peterh : You misunderstand the definition of "homework" on this site. Full title is "homework and exercises," which includes all questions of a "homework-like" nature, even if done for self-study. See meta.physics.stackexchange.com/questions/714 $\endgroup$ – sammy gerbil Aug 4 '16 at 9:45
  • $\begingroup$ @sammygerbil You are right. Although I think, maybe the homework (& exercises) policy could be softened on pragmatical reasons, ref1, ref2. $\endgroup$ – user259412 Aug 4 '16 at 15:57
  • $\begingroup$ @peterh : I agree, the policy should be revised, although I think it should not depend on whether the question has actually been set for homework - that would be too difficult to verify. I agree this question should be on topic. But according to the official policy I think it is not. If you want to change policy, get involved on the Physics Meta site. $\endgroup$ – sammy gerbil Aug 4 '16 at 16:19

First, the reason the child is moving at all is that the potential energy at the top is converted into kinetic energy at the bottom. If the initial height is $h$ and the mass of the child is $m$, we have a final speed $v$ of $$mgh=\frac{1}{2}mv^2\to v=\sqrt{2gh}$$ This is clearly independent of the path the child takes to get to the bottom, and so given that $h$ is the same in all cases, $v$ will also be the same. The child could be moving at a non-zero initial velocity, but this doesn't affect the result; the initial energy (and thus the final energy) will still be the same.

Also, the direction of velocity (a vector), though not the magnitude, depends on the angle of the slide; speed (a scalar) does not. Be careful that you don't confuse the two. The velocities are different; the speeds are the same.

As Floris pointed out, friction could make a different if it existed, but we're assuming that the slides are ideal and frictionless. Friction depends on the normal force exerted on the child, which does depend on the angle of the slide; the work done by friction would reduce the energy. However, we don't have to worry about this here.

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    $\begingroup$ The key here is that it is a frictionless slide. That deserves emphasizing. Otherwise no quibbles, good answer. $\endgroup$ – Floris Aug 4 '16 at 2:49
  • $\begingroup$ @Floris Good point; I added that. $\endgroup$ – HDE 226868 Aug 4 '16 at 2:52
  • $\begingroup$ @Floris Actually - surprisingly! - even if the slide has friction, the velocity at the bottom is the same in each case. $\endgroup$ – sammy gerbil Aug 4 '16 at 16:20
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    $\begingroup$ @sammygerbil only if the horizontal distance is the same - which it doesn't look like it is. $\endgroup$ – Floris Aug 4 '16 at 16:21
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    $\begingroup$ @Floris : Ah yes, I couldn't remember exactly what the catch is! Of course, a vertical drop does no work against friction. A, C, D look the same, but not B. $\endgroup$ – sammy gerbil Aug 4 '16 at 16:26

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