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kikzen
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#1
Report Thread starter 16 years ago
#1
Need some help!

Edexcel M2 ex 3C p74 q20

20. A rough plane is inclined at arcsin 3/5 to the horizontal. A package of mass 40kg is released from rest and travels 15m while increasing speed to 12ms^-1. Find, using energy considerations, the coefficient of friction between the package and the plane.


Answer = 0.138


now what i thought was

Work done against friction = energy lost/gained
fs = pe lost - ke gained
FS = mgh - 1/2mv^2

f = mu(R)
R = 40g4/5
f = mu(40g4/5)

mu(40g4/5)(15) = 3528 - 2880

rearrange and mu = 2.something

i tried anotherway and got 0.6... any ideas ?
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GeneralGrievous
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Report 16 years ago
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energy at top = energy at bottom + work done

40*9.81*(3/5*15) = 1/2*40*12*12 + 15*F

=> F = 43.44 N

μ = F/R = 43.44/(40*9.8*4/5) = 0.138
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kikzen
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Report Thread starter 16 years ago
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(Original post by GeneralGrievous)
energy at top = energy at bottom + work done

40*9.81*(3/5*15) = 1/2*40*12*12 + 15*F

=> F = 43.44 N

μ = F/R = 43.44/(40*9.8*4/5) = 0.138
i could kiss you, man.

but instead ive given you some rep.

cheers

*******s, looking back on what i did - i did it right; its just that i forgot to divide by the 15.... damn!
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