A) 0.8
B) 0.4
C) 0.2
D) 0.1
Correct Answer: A
Solution :
Here: Mass of the car \[m=1500\text{ }kg\] Speed of the car \[v=12.5\text{ }m/s\] Radius of circular path \[r=20\text{ }m\] Then the relation for centripetal force, we have \[F=\frac{m{{\upsilon }^{2}}}{r}=\frac{1500\times {{(12.5)}^{2}}}{20}\] \[=1.172\times {{10}^{4}}N\] So, if the car is not to slip, their frictional force must be less of equal to the centripetal force. Hence, the coefficient of friction between the car and road is given by \[\mu =\frac{F}{mg}\] \[=\frac{1.172\times {{10}^{4}}}{1500\times 9.8}=0.8\]
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