Question

Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its initial speed as vo= 35m/s and the static coefficient to be .50. Assume that the car's weight is distributed evenly on the four wheels, even during braking.
a.) what magnitude of static friction is needed between the tires and road to stop the car just as it reaches the wall?
b.) what is the max possible static friction?
c.) if the coefficient of kinetic friction betweeen the sliding tires and the road is .40, at what speed will the car hit the wall.
d.) to avoid the crash, a driver could elect to turn the car so that it just barely misses the wall. What magnitude of frictional force would be required to keep the car in a circular path of radius d and at the given speed vo?
e.) is the required force that the maximum static friction so that a circular path is possible?

Answer 1

Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

Initial Kinetic energy

= .5 x 1400 x 35 x 35

= 857500 J

Kinetic energy at the at of collision

= 857500 - 587216

= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

= μmg

= .5 x 1400 x 9.8

= 6860 N

So this is not possible.