Suppose an object is sliding w/ a speed less than escape velocity, on an infinite frictionless plane in contact with the surface of Earth. Describe its motion. (Will it slide forerve at a constant velocity? Will it slide to a stop? In what way will its energy changes be similar tothat of a pendulum?)
2006-11-11 02:15:25 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Physics
While you don't have any surface friction, there is still the fluid friction of drag (you didn't specify a vacuum so I must assume there is an atmosphere). Drag is a function based on the drag coefficient and velocity of the object. Since the drag coefficient is constant, and velocity will always be possitive, the object will continue to deccelerate or slow down indefinitely. No matter how slow the object gets, it will never stop entirely because drag decreases with speed.
The force of drag is calculated to this formula F(d) = -1/2*p(v^2)C, where p is density of the fluid, v is velocity of the object and C is the drag coefficient. So, drag increases and decreases exponentially as velocity increases and decreases, with a limit of zero. The velocity will infinitely get closer to zero, but never reach it.
This is similar to a frictionless pendulum in the regards that while the amplitude of the pendulum swings is constantly decreasing due to drag, it will never stop entirely.
2006-11-11 07:24:08 · answer #1 · answered by Spaghetti Cat 5 · 0⤊ 0⤋
Why dont you check Google before asking here?
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2006-11-11 02:19:55 · answer #2 · answered by god knows and sees else Yahoo 6 · 0⤊ 0⤋
You got a lot of bad answers. MikeY's was good. And the one after was OK, but I think we're to assume no wind resistance too.
About a frictionless, dragless, clock-style pendulum. It swings from the lowest position to its highest position (in the path of an arc) where it comes to a stop (because it ran out of kinetic energy) and falls back down.
Your object on a plane also gains altitude as it goes out on the plane. The plane may be at ground level in New York, but in London it would be way overhead. Since it isn't given escape velocity, at some point it will convert all its kinetic energy to potential and stop and reverse direction - falling back on a path that reduces its altitude.
2006-11-11 09:02:15 · answer #3 · answered by sojsail 7 · 0⤊ 0⤋
It's a plane and the earth isn't. As it slides away from the contact point it is actually getting higher so it exchanges kinetic energy for potential energy in the earths gravitational field. Once it's got as high as it can it slides back towards the lowest potential energy point, picking up kinetic energy as it goes ,out-the other side etc,etc.etc. this is why it is just like a pendulum.
2006-11-11 02:40:27 · answer #4 · answered by Anonymous · 1⤊ 0⤋
If it was an infinite flat surface, it would go on forever in a straight line. This is according to newtons law of motion.
If the infinite frictionless plane is a sphere, then it will be as if it is in the lowest orbit possible. If for example it is going directly north, it will go in a straight line over the north and south pole as the revolving shpere rotates underneath it. This will be like an orbit.
2006-11-11 02:29:47 · answer #5 · answered by eric l 6 · 0⤊ 0⤋
all depens if it is fricton less then the object should not stop but then i have to consider that gravity could stop it so i would say that it might take a while but unless it was a friction less bowle that gravity would eventaly stop the object and so long as gravity was below it it should stop dead but on earth the moon would cause the object to move in what way in not sure more then likly like the sea tides up and down so all because your doing this on the earth where you have the moon it would move but then again not meany places in the universe where there is only 1 source of gravity
2006-11-11 02:23:43 · answer #6 · answered by thekonassure 1 · 0⤊ 0⤋
i dont know sorry
2006-11-11 02:18:07 · answer #7 · answered by magdamay34 2 · 0⤊ 2⤋
man do i feel stuiped....lol
2006-11-11 02:19:23 · answer #8 · answered by missy 3 · 0⤊ 2⤋