Meg N.

asked • 10/08/14

Friction and final speed

So the angle is 14°, a=1.42m/s2, Vf=47.7m/s, coeffiecient friction=0.1 and the hill is 800 meters.....
 
Consider how the work done by friction will reduce the final speed
 
 
I set up an equation, but don't know how to solve for the displacement.      v= (square root of) 2(9.8m/s2)(194m)-0.1(9.8m/s2)cos14(d)
 
 
 

1 Expert Answer

By:

Francisco P. answered • 10/08/14

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Meg N.

It asks to consider how the work done by friction will reduce the final speed.  The final speed in the problem without friction ended up being 61.7m/s.  
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10/09/14

Francisco P.

Then the net work would equal the change in kinetic energy: Wnet = ΔK
 
 
Wg + Wfriction = mgh - Ffrictiond = mgh - μmgdcos(θ) = Kf - Ki
 
 
mgh - μmgdcos(θ) = ½mvf2 - ½mvi2
 
 
vf2 = vi2 + 2gh - 2μgdcos(θ)
 
 
vf = √[vi2 + 2gh - 2μgdcos(θ)]
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10/09/14

Francisco P.

 
Looks like you were just missing a factor of 2 in the frictional part of your solution.
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10/09/14

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