Question

4. The tires on a 500 kg race car have a co-efficient of Static Friction (11) of 1.8. (We can use static
friction while tires are rolling, i.e. not moving relative to the ground. If the car skids, then we
would use kinetic friction)
a. What is the maximum braking force (Static Friction) that can be applied to the car?(8820 N)
b. What is the acceleration of the car during braking? (17.64m/s)
c. If the race car has wings which increase the downward force by 4000 N, what
is the maximum braking force? What is the acceleration of the car during braking?(16020 N, 32.04 m/s)
d. Return to the situation of the race car without wings. What is the braking force during
skidding (i.e. kinetic frictional force) if the co-efficient of Kinetic Friction (IX) = 1.2? What is the acceleration
of the car during skidding?(5880 N, 11.76 m/s")
(Does this tell you why drivers avoid skidding to decrease stopping distance?)​

Answer 1

Answer:

braking force =. mass × retardation

accerlation= velocity /time

Answer 2

a. The maximum braking force that can be applied to the car is:

F= μN ( μ= Coefficient of static friction and N= max normal reaction of the car)

= 1.8 X 500g

= 1.8 X 500 X 9.8

= 8820 N

b. Acceleration of the car during braking is:

F=ma (m=mass of the car, a = acceleration)

or 500 X a= 8820

or a = 17.64m/s²

c. Maximum braking force in new situation will be:

F'= μN'

Normal reaction now increases due to the downward force

F'= 1.8 X (500g + 4000)

= 1.8 X 8900

=16020 N

New acceleration a' = F'/m = 32.04 m/s²

d. Braking force during  skidding is:

F''= μ' N (μ' = co-efficient of Kinetic Friction)

F''= 1.2 X 500g

= 5880 N

Acceleration a'' = F''/m = 11.76m/s²

• To avoid skidding, drivers usually slow down their vehicles. This increases the stopping distance.