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A block slides down an inclined surface ...

A block slides down an inclined surface of inclination `30^0` with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the coefficietn of kinetic friction between the two.

Text Solution

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The correct Answer is:
A
From the free body diagram
`R-mg costheta=0`
`rarr R=mg cos theta`
For the block `u=9, S=8m, t=2sec,
From `S=ut+1/2at^2`
`8=0+1/2a2^2`
`rarr a=4m/s^2 `
Again`muR=ma-mg sinthetas=0`
Again `mu R+ma-mgsintheta=0`

From equation i. `mu mg costheta+mamg sintheta` ltbrge0`
`rarr m(mug cos theta+a-gsin theta)=0`
`rarr muxx10xxcos30^0=g sin 30^0-a`
`rarr mu xx 10 sqr3/2=10xx(1/2)-4`
`rarr(5sqrt3)mu=1`
`mu=(1/(5sqt(3)))=0.11`
Hence co efficient of kinetic friction between the two is 0.11.
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