Equal Speeds at y = 0 for Frictionless Slides?

AI Thread Summary
All four frictionless slides described by the equations y = sqrt(x), y = x, y = 2x, and y = x^2 result in equal speeds at y = 0 when starting from the same height h. This conclusion is based on the conservation of mechanical energy, where the initial potential energy converts entirely into kinetic energy at the bottom. Despite differences in the shapes of the slides, the initial potential energy is the same for all, leading to identical final kinetic energies. The confusion about speed arises from misunderstanding that speed is not merely distance over time but also influenced by acceleration. Ultimately, since they all start from the same height, they achieve the same speed at y = 0.
abpandanguyen
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Homework Statement



Consider 4 frictionless slides described by the equations

(1) y = sqrt(x), (2) y = x, (3) y = 2x, (4) y = x2. If you start at rest at y = h and slide down to y = 0, which statement regarding your speed v at y = 0 is correct?

Homework Equations



(1/2)mv2 = KE
mgh = PE

The Attempt at a Solution


The answer to this question is that all the speeds are equal at y = 0

I know the question asks for speed and that speed is distance over time, so doesn't that mean their speeds should differ taking 1, 2, 3, and 4 into account?
 
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Speed is "distance over time" only if there no acceleration, i.e. when the object covers equal distances in equal times. I don't know what y = x2 means, but if it describes a slide, there must be acceleration in case (4) also. Try conservaion of mechanical energy and see what that tells you.
 
o whoops I meant to put that as x^2

I solved for v using conservation of mechanical energy but all my v values came out rather different still. I think I might be doing it wrong...
 
Can you show what you did and exactly how your v values came out wrong?
 
here's one example

1.
mgsqrt(x) = (1/2)mv2

sqrt(2g(sqrt(x))) = v

2.
mgx = (1/2)mv2
sqrt(2gx) = v
 
In all cases the initial kinetic energy is zero and the final potential energy is zero. So the initial potential energy is equal to the final kinetic energy. That's conservation of mechanical energy. Now answer this question, if in all cases the mass starts at height h, what is its initial potential energy?
 
I think I am getting confused by the
y=sqrt(x), y = x etc.

so having that aside, if they all start from the same height, that means they should all have the same initial potential energy and thus same final kinetic energy and given this, when you write the equations out, they all have the same velocity?

is x representing the horizontal distance here?
 
abpandanguyen said:
I think I am getting confused by the
y=sqrt(x), y = x etc.

so having that aside, if they all start from the same height, that means they should all have the same initial potential energy and thus same final kinetic energy and given this, when you write the equations out, they all have the same velocity?

Yes, they have the same speed.

is x representing the horizontal distance here?

Yes, it does.
 
thank you!
want to help me on my other recent post? >_<
 
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