Question

A car of mass 1490 kg makes a 23.0 m radius turn at 7.85 m/s on flat ground. What is the (minimum) coefficient of static friction?

Answer 1

The required centripetal force is given by friction

so

$\frac{mv^{2} }{r}$ = μmg

putting values

$\frac{(1490)(7.85^{2}) }{23}$= μ(1490)(9.8)

on solving

μ = 0.27

hope it helps

Answer 2

Answer:

The answer is 0.273

Explanation:

Let's list everything we know:

m : 1490 kg

r : 23.0 m

v : 7.85m/s

Now using the equation from the lesson, F = mv^2 / r  we can get the centripetal force: 3992 N

Now if you remember from a couple lessons ago, the force is the same as static friction. so we can use the FUN equation... we just have to find the normal, which is 1490 * 9.8 = 14602 N

So, 3992 = mew * 14602, or 3992 / 14602 = mew = 0.273