A) 0.71
B) 0.61
C) 0.51
D) 0.81
Correct Answer: B
Solution :
Key Idea: Body performing circular motion is acted upon by a force directed towards its centre provided by force of friction in our case. When the cyclist moves with a uniform speed v on a circular path of radius r, it has centripetal acceleration\[\left( \frac{{{v}^{2}}}{r} \right),\] hence cyclist is acted upon by a centripetal force directed towards its centre. \[F=\frac{m{{v}^{2}}}{r}\]where m is mass. Also, from Newton's laws \[F=\mu mg\]where \[\mu \]is coefficient of friction. \[\therefore \] \[\frac{m{{v}^{2}}}{r}=\mu mg\]\[\Rightarrow \] \[\mu =\frac{{{v}^{2}}}{rg}\] Putting the numerical values from the question, we have \[\mu =\frac{{{(4.9)}^{2}}}{4\times 9.8}=0.61\]You need to login to perform this action.
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