Question

When you take your 1900-kg car out for a spin, you go around a corner of radius 53 m with a speed of 13 m/s. The coefficient of static friction between the car and the road is 0.88. Assuming your car doesn't skid, (answered by students) What is the force exerted on it by static friction?

Answer 1

Answer:

F = 6058.49 N

Explanation:

given,                                

mass of the car = 1900 Kg

radius of the corner = 53 m

speed of the car = 13 m/s                  

coefficient of static friction = 0.88  

Force exerted = ?                    

$F = \dfrac{mv^2}{r}$            

on balancing car will not skid

$F = \dfrac{1900\times 13^2}{53}$          

F = 6058.49 N                                              

hence, force exerted by the static friction is equal to F = 6058.49 N