A) 0.6 and 0.6
B) 0.6 and 0.5
C) 0.5 and 0.6
D) 0.4 and 0.3
Correct Answer: B
Solution :
Given a plank with a box on it one end is gradually raised about the end having angle of Inclination, the box starts to slip and slides down 4 m the Plank m 4s as shown in figure, The coefficient of static friction, \[{{\mu }_{s}}=\tan {{30}^{{}^\circ }}=\frac{1}{\sqrt{3}}=0.6\] So, distance covered by a plank, \[s=ut+\frac{1}{2}a{{t}^{2}}\] Here, u= 0 and \[a=g\,(\sin \theta -\mu \,\cos \theta )\] \[\therefore \] \[4=\frac{1}{2}g(sin30-{{\mu }_{k}}cos30){{(4)}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,0.5=10\times \frac{1}{2}-{{\mu }_{x}}\times 10\times \frac{\sqrt{3}}{2}\] \[\Rightarrow \,\,\,5\sqrt{3}{{\mu }_{k}}=45\Rightarrow {{\mu }_{k}}=0.51\] Thus, coefficient of kinetic friction between the box and the plank is 0.51You need to login to perform this action.
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