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A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30°, the box starts to slip and slides 4.0m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be, respectively

 

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a
0.5 and 0.6
b
0.4 and 0.3
c
0.6 and 0.6
d
0.6 and 0.5

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detailed solution

Correct option is D

Let μs and μk be the coefficients of static and kinetic friction between the box and the plank respectively. When the angle of inclination θ reaches 30°,  the block just slides,∴  μs=tanθ=tan30°=13=0.6If α is the acceleration produced in the block, thenma=mgsinθ-fk (where fk is force of kinetic friction) =mgsinθ-μkN                             as fk=μkN=mgsinθ-μkmgcosθ                       (as N=mgcosθa=gsinθ-μkcosθ As g=10 ms-2and θ=30°∴  a=10 ms-2sin30°-μkcos30°If  s  is the distance travelled by the block in time t,  then s=12at2   (as u=0 or a=2st2But  s = 4.0 m and  t = 4.0 s (given)∴  a=2(4.0 m)(4.0 s)2=12 ms-2Substituting this value of α in eqn. (i), we get 12 ms-2=10 ms-212-μk32110=1-3μk or 3μk=1-110=910=0.9μk=0.93=0.5

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