A) \[40m{{s}^{-1}}\]
B) \[20m{{s}^{-1}}\]
C) \[15m{{s}^{-1}}\]
D) \[10m{{s}^{-1}}\]
Correct Answer: D
Solution :
Key Idea Force of friction provides the necessary centripetal force. For turning around a curve, centripetal force is provided by friction, so \[\frac{m{{v}^{2}}}{r}<f\] \[\frac{m{{v}^{2}}}{r}<\mu mg\] \[\Rightarrow \] \[v<\sqrt{\mu gr}\] SO that \[{{v}_{\max }}=\sqrt{\mu gr}\] Substituting, \[\mu =0.25,\] \[r=40m,\] \[g=10m{{s}^{-2}},\] we get \[{{v}_{\max }}=10\,m{{s}^{-1}}\].
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