A $100 k g$ 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 $270 N$
$F_{C} = f_{(m a x)} = \frac{m (V_{m a x})^{2}}{R}$
$f_{(m a x)} = \frac{100 \times (9)^{2}}{30}$
$= 270 N$
Option $(C)$

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

The correct option is C 270NFC=f(max)=m(Vmax)2Rf(max)=100×(9)230=270 NOption (C)

A $100 k g$ 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 $270 N$
$F_{C} = f_{(m a x)} = \frac{m (V_{m a x})^{2}}{R}$
$f_{(m a x)} = \frac{100 \times (9)^{2}}{30}$
$= 270 N$
Option $(C)$