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An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m from the barrier. What minimum coefficient of kinetic friction is required to stop the automobile before it hits the barrier?

Answer :

Answer:

μ = 0.408

Explanation:

given,

speed of the automobile (u)= 20 m/s

distance = 50 m

final velocity  (v) = 0 m/s

kinetic friction = ?

we know that,

v² = u² + 2 a s

0 = 20² + 2 × a × 50

[tex]a = \dfrac{400}{2\times 50}[/tex]

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

[tex]\mu = \dfrac{a}{g}[/tex]

[tex]\mu = \dfrac{4}{9.81}[/tex]

μ = 0.408

hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408

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