A) 0.8
B) 0.6
C) 0.4
D) 0.2
Correct Answer: A
Solution :
Mass of the car, m = 1500 kg Speed, \[v=12.5\,m{{s}^{-1}}\] Radius of the circular path = 20 m The centripetal force is given by \[F=\frac{m{{v}^{2}}}{r}=\frac{1500\times {{(12.5)}^{2}}}{20}\] \[=1.172\times {{10}^{4}}N\] Now, if the car does not slip, then the frictional force should be less or equal to the centripetal force. Hence, the coefficient of friction between the tyre and the road is given by \[\mu =\frac{F}{mg}\] \[=\frac{1.172\times {{10}^{4}}}{1500\times 9.8}\] \[=0.8\]
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