GREPhysics.NET: GR9677 #31

GR9677 #31

Problem

GREPhysics.NET Official Solution

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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.

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Comments

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.

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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.

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$\int_0^\infty$	$\int_0^\infty$
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$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

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