Question

three 1500 kg cars are driving at 25m/s along the road. Car A is on a flat portion of the road, car B is at the bottom of a hill and car C is at the top of a hill. (a)Determine the normal force that the road exerts on each car. Suppose each car suddenly brakes hard and starts to skid. (b) Determine the magnitude and direction of each car's acceleration. The coefficient of kinetic friction between the tires and the road is .750

Answer 1

Givenn data:

mass of each car m=1500Kg

speed of each car v=25m/s

a)normal force on car A

Normal force=m*g=1500*9.8=14700N

normal force on car B

N will be enhanced by following the shape of the bottom of the hill

N=m*g+(1/2)m*v²

N=1500*9.8+(1/2)*1500*25²=14700+468750=483450N

normal force on car C

N will be decreased by following the shape of the top of the hill

N=m*g-(1/2)m*v²

N=1500*9.8-(1/2)*1500*25²=14700-468750=-454050N

b)Acceleration of each car is parallel to the road

for car A

F_friction=m*a

F_friction=(coefficient of kinetic friction)(Normal force)

m*a=(coefficient of kinetic friction)(Normal force)

1500*a=(0.750)(14700)

a=7.35m/s²

for car B

F_friction=m*a

F_friction=(coefficient of kinetic friction)(Normal force)

m*a=(coefficient of kinetic friction)(Normal force)

1500*a=(0.750)(483450)

a=241.7m/s²

for car C

F_friction=m*a

F_friction=(coefficient of kinetic friction)(Normal force)

m*a=(coefficient of kinetic friction)(Normal force)

1500*a=(0.750)(-454050)

a=-227m/s²