Question

A car drives around a curve with radius 400 m at a speed of 32 m/s. The road is banked at 7.0 degree. The mass of the car is 1500 kg. What is the magnitude of the centripetal force in order to make this turn

Answer 1

Answer:

The magnitude of the centripetal force to make the turn is 3,840 N.

Explanation:

Given;

radius of the cured road, r = 400 m

speed of the car, v = 32 m/s

mass of the car, m = 1500 kg

The magnitude of the centripetal force to make the turn is given as;

$F_c = \frac{mv^2}{r}$

where;

Fc is the centripetal force

$F_c = \frac{mv^2}{r} \\\\F_c = \frac{(1500)(32)^2}{400}\\\\F_c = 3,840 \ N$

Therefore, the magnitude of the centripetal force to make the turn is 3,840 N.