Question

A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of the resultant force on the 80-kg driver of this car? O a. 0.72 kN O b.0.68 kN O c. 0.64 kN O d.0.76 kN O e. 0.52 kN

Answer 1

c)
`F_(net) = 0.640 kN`

Step-by-step explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

`F_(net) = ma`

here we know

m = 80 kg

for circular motion acceleration is given as

`a_c = (v^2)/(R)`

`a_c = (40^2)/(200) = 8 m/s^2`

now we have

`F_(net) = 80 * 8`

`F_(net) = 640 N`

`F_(net) = 0.640 kN`