GR9677 #31



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Comments 
Ryry013 20190924 06:27:18  For choosing between D and E, you can imagine dropping a feather vs a ball in air. A feather goes slowly, a ball goes quickly, so mass matters. Choose E. (See the other user answers for how to eliminate AC).   Skribb 20090926 04:57:38  (A) Since the sphere starts at rest the kinetic energy must increase as the velocity increases to its terminal velocity.
(B) The sphere reaches a terminal velocity due to the retarding force but the retarding force doesn't completely stop the sphere, so the kinetic energy doesn't go to zero.
(C) It doesn't make physical sense for the sphere to exceed the terminal velocity and to then return back down to it, else it wouldn't be the terminal velocity.
(D) The description finally makes sense but we need to figure out what the velocity is dependent on. Set up your force diagram with the force of gravity pointing down and the retarding force pointing up, . Solve for v and you get so v must be dependent on its mass, m, and the and the constant b.
(E) As per the explanation in D, v is dependent on b and m so this is the correct answer   jmason86 20090921 20:19:44  (A) (B) and (C) were all pretty easy to eliminate because they just don't make any physical sense.
The difference between (D) and (E), then, is just whether or not your speed will depend on your mass at all. Think limits: if you have 0 mass, then you have no surface area to drag with.. but I think this basically the buoyant force that the problem statement declares negligible. Hmpf.   dumbguy 20071016 11:30:43  Just think of it like a free falling object in which air is the vicous medium. What happens to a free falling object then will happen to this sphere, which leaves you at E.   keflavich 20051111 11:15:19  You can also recall terminal velocity happens when , i.e. when the acceleration is zero, which clearly shows that v depends on b and m.
u0455225 20080622 14:32:03 
This implies that equation given in part (D) has a sign error. mgbv=0 when a=0, but the equation in part (D) contends (perhaps unintentionally) that mg+bv=0 when a=0.

physicsisgod 20081028 15:53:26 
u0455225, you are right. Which term is positive or negative depends on how you define the yaxis, but I would probably write , where acceleration is positive in the negativey direction. It doesn't really matter though, as long as they're opposite signs.

 




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