question_answer

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 \[30{}^\circ \], the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be, respectively (Take\[g=10\,\text{m}\,{{\text{s}}^{-2}}\])

A) 0.5 and 0.6

B) 0.4 and 0.3

C) 0.6 and 0.6

D) 0.6 and 0.5

Correct Answer: D

Solution :

[d]: Let \[{{\mu }_{s}}\]and\[{{\mu }_{k}}\]be the coefficients of static and kinetic friction between the box and the plank respectively. When the angle of inclination \[\theta \]reaches \[30{}^\circ \], the block just slides, \[\therefore \]\[{{\mu }_{s}}=\tan \theta =\tan {{30}^{o}}=\frac{1}{\sqrt{5}}=0.6\] If a is the acceleration produced in the block,       then                           \[ma=mg\sin \theta -{{f}_{k}}\] \[=mg\sin \theta -{{\mu }_{k}}N\] (where \[{{f}_{k}}\]is force of kinetic friction as \[{{f}_{k}}={{\mu }_{k}}N\]) \[a=g(sin\theta -{{\mu }_{k}}\cos \theta )\]\[(as\,N\,=mg\cos \theta )\] As\[g=10\,m{{s}^{-2}}\]and\[\theta ={{30}^{o}}\]      \[\therefore \]\[a=(10\,m\,{{s}^{-2}})(sin{{30}^{o}}-{{\mu }_{k}}cos{{30}^{o}})\]      ...(i) If s is the distance travelled by the block in time t, then \[s=\frac{1}{2}a{{t}^{2}}(as\,u=0)or\,a=\frac{2s}{{{t}^{2}}}\] But 5 = 4.0 m and t == 4.0 s (given) \[\therefore \]\[a=\frac{2(4.0m)}{{{(4.0s)}^{2}}}=\frac{1}{2}m{{s}^{-2}}\] Substituting this value of a in equation (i), we get \[\frac{1}{2}m{{s}^{-2}}=(10m{{s}^{-2}})\left( \frac{1}{2}-{{\mu }_{k}}\frac{\sqrt{3}}{2} \right)\] \[\frac{1}{10}=1-\sqrt{3}{{\mu }_{k}}\]or\[\sqrt{3}{{\mu }_{k}}=1-\frac{1}{10}=\frac{9}{10}=0.9\] \[{{\mu }_{k}}=\frac{0.9}{\sqrt{3}}=0.5\]