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\[\sqrt{3}{{\mu }_{k}}=1-\frac{1}{10}=\frac{9}{10}=0.9\] \[{{\mu }_{k}}=\frac{0.9}{\sqrt{3}}=0.5\]You need to login to perform this action.
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