Question

A car is moving on a horizontal circular track of radius 0.2km with a constant speed.if coefficient of friction between tyres of car and road is 0.45,then speed of car may be (g=10m/s^2)

Answer 1

Since car is moving on rough surface

The force of friction on the car will provide centripetal force to make the turning of car

So we can write

$F_c = \frac{mv^2}{R}$

here

$F_c$ = Net force towards the center of the path

m = mass of the car

v = speed of the car

r = radius of circular path

So here we can say that towards the center of the path only one force is acting that is friction force

so we can write

$F_c = F_f = \mu m g$

$\mu mg = \frac{mv^2}{r}$

now by rearranging the terms

$v = \sqrt{\mu r g}$

now plug in all data in above equation

$v = \sqrt{0.45 * 200 * 10}$

$v = 30 m/s$

so the speed of car must be 30 m/s.

Answer 2

Answer:

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