GR8677 #93


Problem



Mechanics}Power
Recall the following basic formulas, , where is power, is current, is voltage, is force, and is velocity. , where is the efficiency, which relates work (and thus power).
The problem gives , , , , where all units are SI.
Thus . Solve for , as in choice (D).


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Comments 
trous. 20131004 04:44:15  F * x=P*t)=>F=P/v=9*120/10=108.
F=u*N)=>u=108/100=1.08   newtonsfourth 20090720 16:16:16  Aluminum in contact with another piece of aluminum has a coefficient of friction greater than 1 so it shears instead of slides. interesting thing to see   Poop Loops 20080923 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.
CaspianXI 20090314 15:10:28 
No, the normal force IS normal to the surface. It's the force of friction that's pushing parallel to the surface.
The force of friction (parallel to the surface) is directly related to the normal force (normal to the surface). An intuitive way to think about this is if you lift the sander off of the surface (normal force = 0), the frictional force goes away. If you push down really hard (normal force increases), the frictional force increases.

  Furious 20071101 18:40:00  Was I the only one who thought that you couldn't have a coefficient of friction greater than one?
shak 20100801 09:23:55 
you are not the only one, i also thought so:))

his dudeness 20100819 18:37:17 
Yeah I definitely fell for it too... Here's Wikipedia to the rescue: "Occasionally it is maintained that µ is always < 1, but this is not true. While in most relevant applications µ < 1, a value above 1 merely implies that the force required to slide an object along the surface is greater than the normal force of the surface on the object. "

 

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