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?

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

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

μ = 0.408

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

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.