When a car of mass $1200 k g$ is moving with a velocity of $15 m s^{- 1}$ on a rough horizontal road. Its engine is switched off. How far does the car travel before it comes to rest if the coefficient of kinetic friction between the road and tyres of the car is $0.5$ ? $(g = 10 m s^{- 2})$ .

6.2 m

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20 m

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22.5 m

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30 m

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Solution

The correct option is A $6.2 m$
Retardation of car = force of friction/ man of car

$= \frac{μ_{k} m g}{m} = μ_{k} g$

$- a = μ_{k} g$

$v^{2} - u^{2} = 2 a s$

$s = \frac{v^{2}}{2 μ_{k} g}$

$s = \frac{20^{2}}{2 \times 0.75 \times 10} = \frac{400}{15} = 26.67 m$

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When a car of mass 1200 kg is moving with a velocity of 15 ms−1 on a rough horizontal road. Its engine is switched off. How far does the car travel before it comes to rest if the coefficient of kinetic friction between the road and tyres of the car is 0.5? (g=10ms−2).

The correct option is A 6.2 mRetardation of car = force of friction/ man of car =μkmgm=μkg −a=μkg v2−u2=2as s=v22μkg s=2022×0.75×10=40015=26.67m

When a car of mass $1200 k g$ is moving with a velocity of $15 m s^{- 1}$ on a rough horizontal road. Its engine is switched off. How far does the car travel before it comes to rest if the coefficient of kinetic friction between the road and tyres of the car is $0.5$ ? $(g = 10 m s^{- 2})$ .

The correct option is A $6.2 m$
Retardation of car = force of friction/ man of car

$= \frac{μ_{k} m g}{m} = μ_{k} g$

$- a = μ_{k} g$

$v^{2} - u^{2} = 2 a s$

$s = \frac{v^{2}}{2 μ_{k} g}$

$s = \frac{20^{2}}{2 \times 0.75 \times 10} = \frac{400}{15} = 26.67 m$