Question

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%.

 

Answer 1

$N= Mg \cos{\alpha} : Normal Reaction\\ F_R=\mu Mg \cos{\alpha}: Friction froce$

Writing equation of motion.......

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

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

$$\mu=\tan\alpha=\tan{23\degree}\\ \mu=0.424..........(a)Ans$$

And Hence

$$F_R=.424\times 6.5\times 9.8\times \cos 23\degree\\ F_R=24.861 N...........(b)Ans$$

Now part (c)

$$\mu\to \cfrac{3}{4}\mu$$

Now again writing equation of motion,

$$M a' = Mg \sin\alpha-\cfrac{3}{4}\mu Mg \cos\alpha\\ a'=g\sin\alpha-\cfrac{3}{4}\mu g \cos\alpha\\ a'=9.8\times \sin23\degree -\cfrac{3}{4}\times 0.424\times9.8\times\cos23\degree\\ a'=3.829-2.869=0.96 meter/sec$$

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