1. Car Down Incline w Friction An automobile weighing 4250 lb starts from rest at point A on a 6o incline and coasts through a distance of 500 ft to point B. The brakes are then applied, causing the automobile to come to a stop at point C, 70 ft from B. Knowing that slipping is impending during the braking period and neglecting air resistance and rolling resistance, determine (a) the speed of the automobile at point B, (b) the coefficient of static friction between the tires and the road.

Explanation:

Let θ be the inclination

downward acceleration on an inclined plane

= g sinθ

= 32 x sin6

a =  3.345 ft /s

a ) for knowing the speed at point B

v² = u² + 2 a s , v is final velocity , u is initial velocity , a is acceleration and s is distance travelled .

v² = 0 + 2 x 3.345 x 500

= 3345

v = 57.8 ft /s

from point B to C , the car decelerates so we shall find deceleration

v² = u² + 2 a s

0 = 3345 + 2 x a x 70      ( v becomes u here )

a = - 23.9 m /s²

net force on car during deceleration

= μmgcosθ - mg sinθ     where  μ is coefficient of static friction ,

= mg ( μcosθ -  sinθ )

deceleration = g ( μcosθ -  sinθ )

g ( μcosθ -  sinθ )  = 23.9

( μcosθ -  sinθ ) = .74

μcosθ = .74 + .104

= .8445

μ = .8445 / .9945

= .85 .

a straight line sloped

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