Answer to Question #149739 in Mechanics | Relativity for Mylene

Question #149739

1. A block is sliding at a constant speed down a wooden board inclined at an angle 23º with the horizontal surfaces. The mass of the block is 6.5 kg. Find the (a) coefficient of kinetic friction and (b) kinetic friction. (c) Explain what would happen if the board is lubricated so that friction is reduced to 75%.


 


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Expert's answer
2020-12-10T11:06:37-0500

N=Mgcosα:NormalReactionFR=μMgcosα:FrictionfroceN= Mg \cos{\alpha} : Normal Reaction\\ F_R=\mu Mg \cos{\alpha}: Friction froce

Writing equation of motion.......

Ma=MgsinαμMgcosαMa=Mg \sin\alpha-\mu Mg \cos\alpha

Where "a" is the acceleration of the block which is zero since block is moving with constant velocity. Hence......

μ=tanα=tan23°μ=0.424..........(a)Ans\mu=\tan\alpha=\tan{23\degree}\\ \mu=0.424..........(a)Ans

And Hence

FR=.424×6.5×9.8×cos23°FR=24.861N...........(b)AnsF_R=.424\times 6.5\times 9.8\times \cos 23\degree\\ F_R=24.861 N...........(b)Ans

Now part (c)

μ34μ\mu\to \cfrac{3}{4}\mu

Now again writing equation of motion,

Ma=Mgsinα34μMgcosαa=gsinα34μgcosαa=9.8×sin23°34×0.424×9.8×cos23°a=3.8292.869=0.96meter/secM a' = Mg \sin\alpha-\cfrac{3}{4}\mu Mg \cos\alpha\\ a'=g\sin\alpha-\cfrac{3}{4}\mu g \cos\alpha\\ a'=9.8\times \sin23\degree -\cfrac{3}{4}\times 0.424\times9.8\times\cos23\degree\\ a'=3.829-2.869=0.96 meter/sec

we see that if board is lubricated then block starts accelerating.

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