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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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||||| A 1.5 kg block and a 2.5 kg block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless
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Answer

The lighter block will have the Positive Acceleration that is +2.45 m/s square.

A (1.5 Kg) = + 2.45 m/s square

Explanation:

To solve this there is a hard way to do this and there is an easy way to do this. The hard way is to solve Newton's second law for each block individually and then combine them and you get two equations with two unknowns. you try your best to solve the algebra without losing any sins but lets be honest it usually goes wrong.

So the easy way to do this the way to get the magnitude of the acceleration of the blocks. That is to say that i want to know the magnitude at which 2.5 Kg block accelerates or 1.5 Kg block accelerates when the the blocks were released.

Take the net external force that tries to make system go and divide it by total mass of the the system.

<u>A </u><u>(of System) = </u><u>F</u><u>(net external force) / </u><u>m</u><u> (total mass of system)</u>

This is the quick way to know the magnitude of acceleration of the objects in the system. But this is only possible if the system is moved in same magnitude of acceleration that is 2.5 Kg block will move downward and 1.5 Kg block will move upward with the same magnitude. So here in this case we have friction-less pulley and the blocks will move with the same magnitude of acceleration.

To find the external forces

External forces are the forces which exerted on the objects in our system from the objects outside of our system. So one external force is the force of gravity. Both 2.5 Kg block and the force of gravity will be in downward direction.

Force of gravity on 2.5 Kg block

F = + (2.5 x 9.8) = 24.5

After releasing the rope the 2.5 Kg block will drive the system and accelerates in downward direction so It will be a positive force.

Force of gravity on 1.5 Kg block

F = - (1.5 x 9.8) = 14.7

The force of gravity on 1.5 Kg block will be negative because it will accelerate in upward or opposite direction of the force of gravity. Because the whole system is moving in one direction but the force of gravity on 1.5 Kg block is opposing the acceleration of the system.

Now divide the the Net external forces by total mass of the blocks that is

<u>A </u><u>(of System) = </u><u>F</u><u>(net external force) / </u><u>m</u><u> (total mass of system)</u>

A = (+ 24.5 - 14.7) / 2.5 + 1.5

A = 9.8 / 4

A = 2.45 meter per second square

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A (1.5 Kg) = - 2.45 m/s square

So the Acceleration of 1.5 Kg block will be Positive that is +2.45 m/s square. Since block is accelerating up and we usually treat up as positive.

A (1.5 Kg) = + 2.45 m/s square

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