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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?
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2655 Upvotes
Verified Answer
The correct answer is:
$0.8$
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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