GREPhysics.NET: GR9677 #31

GR9677 #31

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Mechanics $\Rightarrow$ }Frictional Force

(A) A falling object experiencing friction falls faster and faster until it reaches a terminal speed. Its kinetic energy increases proportional to the square of the velocity and approaches a asymptotic value.

(B) The kinetic energy increases to a maximum, but it does not decrease to 0. See (A).

(C) The maximal speed is the terminal speed.

(D) One has the equation $m\ddot{y}+b\dot{y}+mg=0\Rightarrow m\dot{v}+bv+mg=0$ . Without having to solve for $v$ , one can tell by inspection that $v(t)$ will depend on both $b$ and $m$ .

(E) See (D). This is the remaining choice, and it's right.

Alternate Solutions

There are no Alternate Solutions for this problem. Be the first to post one!

Comments

Ryry013
2019-09-24 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 A-C).

Reply to this comment

Skribb
2009-09-26 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, $F_{net}=ma=bv-mg$ . Solve for v and you get $v=\frac{ma+mg}{b}$ 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

Reply to this comment

jmason86
2009-09-21 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.

Reply to this comment

dumbguy
2007-10-16 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.

Reply to this comment

keflavich
2005-11-11 11:15:19

You can also recall terminal velocity happens when $bv=mg$ , i.e. when the acceleration is zero, which clearly shows that v depends on b and m.

u0455225
2008-06-22 14:32:03

This implies that equation given in part (D) has a sign error. mg-bv=0 when a=0, but the equation in part (D) contends (perhaps unintentionally) that mg+bv=0 when a=0.

physicsisgod
2008-10-28 15:53:26

u0455225, you are right. Which term is positive or negative depends on how you define the y-axis, but I would probably write $\ F=m \ddot{x} = -k \dot{x} + mg$ , where acceleration is positive in the negative-y direction. It doesn't really matter though, as long as they're opposite signs.

Reply to this comment

Post A Comment!

You are replying to:

(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, $F_{net}=ma=bv-mg$ . Solve for v and you get $v=\frac{ma+mg}{b}$ 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

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone