Question

If the coefficient of static friction between your friend and the car seat is 0.500 and you keep driving at a constant speed of 21.0 m/sm/s , what is the maximum radius you could make your turn and still have your friend slide your way?

Answer 1

Answer:

90m

Explanation:

Let g = 9.8 m/s2. The friction is the product of normal force and its coefficient, and normal force is equal to gravity

$F_f = \mu N = \mu mg$

The acceleration caused by friction, according to Newton's 2nd law:

$a_f = F_f / m = \mu g = 0.5 * 9.8 = 4.9 m/s^2$

For the friend to slide over, then the centripetal acceleration must be equal to friction acceleration.

Since you are driving at a constant speed of 21 m/s, then your maximum radius of curvature can be calculated using the following formula:

$a_c = a_f = v^2/r = 4.9$

$r = v^2/4.9 = 21^2/4.9 = 90 m$