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Serggg [28]
3 years ago
5

there are two slides at the park between which you are deciding. Both start at the height of 6 meter. One is short and Steve whi

le the other is long and shallow. If both sides are frictionless, which one should you choose if you want to be moving as fast as possible at the bottom? A. short and steep. B. long and shallow. C. it doesn't matter.
Physics
1 answer:
kumpel [21]3 years ago
6 0

It doesn't matter.  If the slides are truly frictionless, then
your kinetic energy at the bottom will be equal to the
potential energy you had at the top, no matter what kind
of route you took getting down.
___________________________

The only way I can think of that it would make a difference
would be if the shallow slide were REALLY REALLY long,
and you didn't have anything to eat all the way down. 
Then you might lose some weight while you're on the slide,
and your mass might be less at the bottom than it was at the
top.  Then, in order to have the same kinetic energy at the
bottom, you'd need to be going a little bit faster.

But if it takes less than, say, two or three days, to go down the
long, shallow slide, then this effect would probably be too small
to make any difference.

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Answer:

a)6.7m/S

b)6.8m/s

Explanation:

Hello ! To solve the point b you must follow the steps below

1.Draw the slide taking into account its length and height and find the angle from which the swimmer is launched (see attached image)

2. Find the horizontal velocity (X) and vertical (Y) components (see attached image)

3) for the third step we must remember that as in the slide there is no horizontal acceleration the speed in X will remain constant at the end of the swimmer's path (Vx = 0.59m / s)

4)

the fourth step is to remember that vertically there is constant acceleration called gravity (g = 9.81m / s ^ 2), so to find the speed at the end of the route we use the following equation

Vfy= \sqrt{Vy^2+2gy}

where    

Vfy= final verticaly speed    

Vy=initial verticaly speed=0.59m/S

g=gravity=9.81m/S^2

y=height of slide=2.31m

solving

Vfy= \sqrt{Vy^2+2gy}\\Vfy= \sqrt{(0.59)^2+2(9.81)(2.31)}=6.77m/s

The last step is to add the velocity components vectorally at the end of the route with the following equation

V=\sqrt{Vfy^2+Vx^2} =\sqrt{6.77^2+0.59^2} =6.8m/s

point A

taking into account the previous steps we can infer that as the swimmer starts from rest, the velocity (Vx=Vy=O) is zero, so we should only use the formula for constant acceleration movement.

Vfy= \sqrt{Vy^2+2gy}

vy=0

Vfy=\sqrt{2gy}

Vfy=\sqrt{2(9.81)(2.31)}=6.7m/s

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