A car driven round curved path of radius $18 m$ without the danger of skidding. The coefficient of friction between the tyres of the car and the surface of the curved path is $0.5$ . What is the maximum speed in kmph of the car for safe driving nearly? ( $g = 10 {m s}^{- 2}$ )

9.3

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34.48

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18.42

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20.15

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Solution

The correct option is B $34.48$
Frictional force= centrifugal force
$f m g = \frac{m v^{2}}{R}$

where,
$f =$ coefficient of friction
$m =$ mass of vehicle $(k g)$

$g =$ acceleration due to gravity $(m / s^{2})$

$v =$ speed of vehicle $(m / s)$

$R =$ radius of curve $(m)$

$f = \frac{v^{2}}{g R}$

$0.5 = \frac{v^{2}}{10 \times 18}$

$v^{2} = 0.5 \times 10 \times 18$

$v = \sqrt{90}$

$v = 9.45 m / s = 9.45 \times \frac{18}{5} = 34.02 k m p h$

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A car driven round curved path of radius 18m without the danger of skidding. The coefficient of friction between the tyres of the car and the surface of the curved path is 0.5. What is the maximum speed in kmph of the car for safe driving nearly? (g=10ms−2)

The correct option is B 34.48Frictional force= centrifugal forcefmg=mv2Rwhere,f=coefficient of frictionm=mass of vehicle (kg)g=acceleration due to gravity (m/s2)v=speed of vehicle (m/s)R=radius of curve (m)f=v2gR0.5=v210×18v2=0.5×10×18v=√90v=9.45m/s=9.45×185=34.02kmph

A car driven round curved path of radius $18 m$ without the danger of skidding. The coefficient of friction between the tyres of the car and the surface of the curved path is $0.5$ . What is the maximum speed in kmph of the car for safe driving nearly? ( $g = 10 {m s}^{- 2}$ )