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

Question

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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