Question

What is the maximum speed with which a 1200-kg car can round a turn of radius 88.0 m on a flat road if the coefficient of static friction between tires and road is 0.40?

Answer 1

Answer:

18.6 m/s

Explanation:

The frictional force acting on the car is given by:

$F_f = \mu m g=(0.40 )(1200 kg)(9.81 m/s^2)=4709 N$

But the frictional force also corresponds to the centripetal force that keeps the car in circular motion in the turn:

$F_f = F_c = m \frac{v^2}{r}$

Where v is the maximum speed the car can achieve remaining in the turn. Substituting r=88.0 m and re-arranging the formula, we can find the value of v:

$v=\sqrt{\frac{Fr}{m}}=\sqrt{\frac{(4709 N)(88.0 m)}{1200 kg}}=18.6 m/s$