A 900 kg car is beings coasting at 20 m/s along a level road . How large of a constant frictional force is required to stop it in a distance of 30m ?

Answer :

Given data

*The given mass of the car is m = 900 kg

*The car is moving at a velocity is u = 20 m/s

*The given distance is s = 30 m

The formula for the acceleration of the car is given by the equation of motion as

[tex]\begin{gathered} v^2=u^2+2as \\ a=\frac{v^2-u^2}{2s} \end{gathered}[/tex]

*Here v = 0 m/s is the final velocity of the car

Substitute the known values in the above expression as

[tex]\begin{gathered} a=\frac{(0)^2-(20)^2}{2\times30} \\ =-6.67m/s^2 \end{gathered}[/tex]

The formula for the frictional force is required to stop it at a distance of 30 m is given as

[tex]f=ma[/tex]

Substitute the known values in the above expression as

[tex]\begin{gathered} f=(900)(-6.67) \\ =-6000\text{ N} \end{gathered}[/tex]

Hence, the frictional force is required to stop it at a distance of 30 m is f = -6000 N

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