Question

To test the performance of its tires, a car

travels along a perfectly flat (no banking) cir-

cular track of radius 516 m. The car increases

its speed at uniform rate of

at ≡

d |v|

dt = 3.89 m/s

2

until the tires start to skid.

If the tires start to skid when the car reaches

a speed of 32.8 m/s, what is the coefficient of

static friction between the tires and the road?​

Answer 1

The coefficient of static friction is 0.213

Explanation:

In order for the car to stay in circular motion, the force of friction acting on the tires must be equal to the centripetal force. Therefore we can write

$\mu mg = m\frac{v^2}{r}$

where

$\mu$ is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

When the car reaches a speed of

v = 32.8 m/s

The tires start to skid: this means that for values of v larger than this value, the force of friction is no longer able to provide the needed centripetal force.

We have the following data:

r = 516 m is the radius of the curve

$g=9.8 m/s^2$

Solving the equation for $\mu$, we find the coefficient of friction:

$\mu = \frac{v^2}{rg}=\frac{(32.8)^2}{(516)(9.8)}=0.213$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly