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

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.