Question

A car of mass $1500kg$ is moving with a speed of $12.5m{s}^{-1}$ on a circular path of radius $20m$ on a level road. The value of coefficient of friction between the tyres and road, so that the car does not slip, is

• A. 0.8
• B. 0.6
• C. 0.4
• D. 0.2

Answer 1

Answer: (A)

Mass of the car, $m=1500kg$
Speed, $v=12.5m{s}^{-1}$
Radius of the circular path $=20m$
The centripetal force is given by
$F=\frac{m{v}^2}{r}=\frac{1500\times (12.5{)}^2}{20}$
$=1.172\times 10^{4}N$
Now, if the car does not slip, then the frictional force should be less or equal to the centripetal force.
Hence, the coefficient of friction between the tyre and the road is given by
$\mu =\frac{F}{mg}$
$=\frac{1.172\times 10^4}{1500\times 9.8}$
$=0.8$