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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Infinity Learn · Accepted Answer

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,∴  μ$$s=$$$$tan$$θ$$=tan30$$°$$=13=0.6If$$ α is the acceleration produced in the block, $$thenma=mgsin$$θ$$-fk$$ (where$$ fk is force of kinetic $$friction) $$=mgsin$$θ$$-$$μkN                             as $$fk=$$μ$$kN=mgsin$$θ$$-$$μkmgcosθ                       (as$$ $$N=mgcos$$θ$$a=gsin$$θ$$-$$μkcosθ As $$g=10$$ $$ms-2and$$ θ$$=30$$°∴  $$a=10$$ $$ms-2sin30$$°$$-$$μ$$kcos30$$°If  s  is the distance travelled by the block in time t,  then $$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 $$12$$ $$ms-2=10$$ $$ms-212-$$μ$$k32110=1-3$$μk or $$3$$μ$$k=1-110=910=0.9$$μ$$k=0.93=0.5$$

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