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Verbatim question for GR8677 #93
Mechanics}Power


Recall the following basic formulas, P=IV=Fv, where P is power, I is current, V is voltage, F is force, and v is velocity. e=\frac{Work_{done}}{Work_{input}}, where e is the efficiency, which relates work (and thus power).

The problem gives e=1, F=\mu N=100\mu, v=10, I=9, V=120---where all units are SI.

Thus e=1 \Rightarrow P_{done}=P_{input} \Rightarrow \mu N v = IV. Solve for \mu=\frac{IV}{Nv}=\frac{9*120}{100*10}=108/100\approx 1.1, as in choice (D).

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Comments
Poop Loops
2008-09-23 21:04:09
Not only did I think that you couldn't have a CoF > 1, since friction is only a retarding force, but I also thought "normal" meant "normal to the surface". I had no idea it was pushing parallel to the surface. I just didn't understand how a vertical force related to the parallel energy being released.

Apparently you CAN have CoF > 1, though.
NEC
Furious
2007-11-01 18:40:00
Was I the only one who thought that you couldn't have a coefficient of friction greater than one?NEC

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