A car of mass 800 kg moves on a circular track of radius 40 m. If the coefficient of friction is 0.5, then maximum velocity with which the car can move is

7 m/s

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14 m/s

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8 m/s

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12 m/s

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Solution

The correct option is B 14 m/s
$v_{m a x} = \sqrt{μ r g} = \sqrt{0.5 \times 40 \times 9.8} = 14 m / s$

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A car of mass 800 kg moves on a circular track of radius 40 m. If the coefficient of friction is 0.5, then maximum velocity with which the car can move is

The correct option is B 14 m/svmax=√μrg=√0.5×40×9.8=14m/s

The correct option is B 14 m/s
$v_{m a x} = \sqrt{μ r g} = \sqrt{0.5 \times 40 \times 9.8} = 14 m / s$