Question

A woman is driving her truck with speed 45.0 mi/h on a horizontal stretch of road. (a) When the road is wet, the coefficient of static friction between the road and the tires is 0.105. Find the minimum stopping distance (in m). m (b) When the road is dry, μs = 0.602. Find the minimum stopping distance (in m).

Answer 1

Answer:

Explanation:

45 mi /h = 45 x 1.6 x 1000 / (60 x 60) m /s

= 20 m /s

Maximum frictional force possible on wet road

= μs x mg where μs is coefficient of static friction and m is mass of body

Applying work energy theorem

work done by friction = kinetic energy of truck

μs x mg x d  = 1/2 m v ² where v is velocity of body and d is stopping distance

2xμs x g x d = v²

2 x .105 x 9.8 x d = 20 x 20

d = 194.36 m

b )

In this case

μs = 0.602

Inserting this value in the relation above

2xμs x g x d = v²

2 x .602 x 9.8 x d = 20 x 20

d = 33.9 m .