Question

A race car travels 44.3 m/s around a banked (45° with the horizontal) circular (radius = 200 m) track. What is the magnitude of the resultant force in kN on the 80 kg driver of this car?

Answer 1

The magnitude of the resultant force is given by the centripetal force, since the car is under a circular motion. So, we have:

$F_c=ma_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Where v is the linear speed and r the radius of the circular motion. Replacing this and solving:

$F=m\frac{v^2}{r}\\F=80kg\frac{(44.3\frac{m}{s})^2}{200m}\\F=785N*\frac{1kN}{1000N}\\F=0.785kN$