Question

A car of mass 1500kg is moving with the speed of 12.5m/s on a circular path of radius 20m on a level road. What should be the value of the coefficient of friction to attain this force?

Pls fast guys..
It's urgent

Answer 1

$<b><i>$

Let μ be the coefficient of friction.

According to question,
$f = \frac{mv^{2}}{r}$ ------[1]

where,
f = frictional force
m = mass of car
r = circular trace

We know,
$f = μN$

& $N= mg$.

So
$f = μmg$ --------[2]

Since both are same forces
[1] = [2]

$\frac{mv^{2}}{r} = μmg$

=> $μ = \frac{v^{2}}{rg}$

Taking g as 10 m/s

Substitute

$r = 20\: m$
$v= 12.5\: m/s$&
$g = 10\: m/s^{2}$

Therefore,
Coefficient of friction
$μ = \frac{(12.5)^{2}}{10×20}$

=> $μ = \frac{156.25}{200}$
=> $μ = 0.78125$

So the value of the coefficient of friction to attain the force is $0.78125$.

✌✌✌

Keep it Mello