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

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