An 1460 kg car passes over a bump in a road that follows the arc of a circle of radius 35.6 m. The acceleration of gravity is 9.8 m/s 2 .

Question

1.What force does the road exert on the car as the car passes the highest point of the bump if the car travels at 14.1 m/s? Answer in units of N.

2.What is the maximum speed the car can have as it passes this highest point before losing contact with the road? Answer in units of m/s.

Answer 1

(1)

The car is on a vertical circular path. At the highest point, the difference between its weight and the upward normal force exerted by the road must equal the centripetal force on the car.

F = upward force of the road = ?

m = mass of car (given)

v = speed of car (given)

r = radius of bump (given)

g = acceleration of gravity

mg - F = mv²/r ⇒

F = mg - mv²/r = m(g - v²/r)

You know m, g, v, and r. Plug in the numbers and get F in newtons.

(2)

u = max speed = ?

At maximum speed, while in contact with the road, the upward normal force must be zero. Use the same equation as in part (1), but set F = 0 and solve for u.

F = m(g - u²/r) = 0 ⇒

u²/r = g

u = √(rg)

Plug in the numbers and get u in m/s. It will come out higher than the v given in part (1).