Question

To test the performance of its tires, a car

Answer 1

Answer:

The coefficient of  static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r =  516 m

Tangential Acceleration $a_r$=  3.89 m/s^2

Speed,v =  32.8 m/s

To Find:

The coefficient of  static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=>$= \sqrt{ (a_r)^2+(a_R)^2}$

=>$\sqrt{ (3.89 )^2+( 2.085)^2}$

=>$\sqrt{ (15.1321)+(4.347)^2}$

=>$19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$