MHT CET · Physics · Laws of Motion
A car of mass \(1500 \mathrm{~kg}\) is moving with a speed of \(12.5 \mathrm{~m} / \mathrm{s}\) on a circular path of radius \(20 \mathrm{~m}\) on a level road. What should be the coefficient of friction between the car and the road, so that the car does not slip?
- A \(0.2\)
- B \(0.4\)
- C \(0.6\)
- D \(0.8\)
Answer & Solution
Correct Answer
(D) \(0.8\)
Step-by-step Solution
Detailed explanation
In this case centripetal force provides by friction
\(\therefore\) \(\frac{m v^{2}}{r}=\mu m g\)
i. \(e\),
\(\mu=\frac{v^{2}}{r g}\)
\(\mu=\frac{12.5 \times 12.5}{20 \times 9.8}=0.8\)
\(\therefore\) \(\frac{m v^{2}}{r}=\mu m g\)
i. \(e\),
\(\mu=\frac{v^{2}}{r g}\)
\(\mu=\frac{12.5 \times 12.5}{20 \times 9.8}=0.8\)
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