A 950 kg car rounds an unbanked curve at a speed of 25 m/s. If the radius of the curve is 72 m, what is the minimum coefficient of friction between the car and the road required so that the car does not skid?

Compute the car's weight:

W = m g = (950 kg) (9.8 m/s²) = 9310 N

The net vertical force on the car is

∑ F = N - W = 0

so the normal force has magnitude

N = W = 9310 N

Then the friction force that keeps the car from skidding has magnitude F = µ N, where µ is the coefficient of friction, and it's friction that makes up the net horizontal force on the car. By Newton's second law, we have

F = m a

µ N = m v ² / R

µ (9310 N) = (950 kg) (25 m/s)² / (72 m)

µ ≈ 0.89

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft