GREPhysics.NET: GR9677 #6

GR9677 #6

Problem

GREPhysics.NET Official Solution

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Mechanics $\Rightarrow$ }Inclined Plane

Set up the usual coordinate system with horizontal axis parallel to incline surface. The equations are, (since the mass slides down at constant speed),

$\begin{eqnarray} \sum F_x &=& 0 = f-mg\sin\theta\\ \sum F_y &=& 0 = N-mg\cos\theta<br /> \end{eqnarray}$

Friction is given by $f=\mu N =\mu mg\cos\theta$ , where the normal force $N$ is determined from the $F_y$ equation. For constant velocity one also has, $f=mg\sin\theta=\mu mg\cos\theta\Rightarrow \mu = \tan \theta$

To find the work done by friction, one calculates $W=f L$ , where $L\sin\theta = h$ . Thus $W = \tan\theta mg \cos\theta \frac{h}{\sin\theta}=mgh$ , as in the almost-too-trivial, but right, choice (B).

Alternate Solutions

99percent
2008-11-06 06:25:06

An easy way...

Since, the speed of the block is constant, there is no gain in the kinetic energy of the block.

Thus, potential energy lost by the block = Energy dissipated by friction = mgh.

Bingo..!!

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99percent
2008-11-06 06:25:06

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cyberdeathreaper
2006-12-28 19:42:46

Isn't it somewhat repetitive to solve for the coefficient of kinetic friction, to plug back in to solve for friction, to calculate work? Couldn't you solve for f from your x-coordinate forces (f=mgsin(theta)), and plug that straight into your work equation (W=fl)?

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daschaich
2005-11-07 23:22:21

There's an easier way to do it - conservation of energy. Since the speed of the block is constant, its kinetic energy is the same at the top and bottom of the ramp. Therefore all its gravitational potential energy (mgh) must have been dissipated by friction.

Blake7
2007-09-22 08:04:31

Excellent observation; saves a tremendous amount of time with much less risk.

dcan
2008-04-09 17:11:02

This seems so obvious after seeing the answer. It's hard to get out of the crunch mode.

tau1777
2008-11-05 14:49:47

all i could say after reading this was:quite beautiful. this is an excellent solution, thank you for sharing.

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