A 100 kg car is moving with a maximum velocity of 9 m/s across a circular track of radius 30 m. The maximum force of friction between the road and the car is

1000 N

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706 N

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270 N

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200 N

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Solution

The correct option is C
270 N

Maximum force of friction = centripetal force
$\frac{m v^{2}}{r} = \frac{100 \times (9)^{2}}{30} = 270 N$

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A 100 kg car is moving with a maximum velocity of 9 m/s across a circular track of radius 30 m. The maximum force of friction between the road and the car is

The correct option is C 270 N Maximum force of friction = centripetal force mv2r=100×(9)230=270 N

The correct option is C
270 N

Maximum force of friction = centripetal force
$\frac{m v^{2}}{r} = \frac{100 \times (9)^{2}}{30} = 270 N$