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# The coefficient of friction between the tires and the road is 0.25. Then find the maximum speed with which a car can be driven around a curve of a radius of 40 m without skidding. (Assume$g = 10\,m{s^{ - 2}}$)A. $40\,m{s^{ - 1}}$ B. $20\,m{s^{ - 1}}$C. $15\,m{s^{ - 1}}$D. $10\,m{s^{ - 1}}$

Last updated date: 12th Aug 2024
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Hint:Frictional force is defined as the opposing force that is created between two surfaces that try to move in the same direction or that try to move in opposite directions. The main purpose of a frictional force is to create the resistance to the motion of one surface over the other surface.

Formula Used:
The formula to find the frictional force between the road and tire is,
$F = \mu mg$
Where, $\mu$ is coefficient of friction, m is mass and g is acceleration due to gravity.

Complete step by step solution:
Then, frictional force between the road and tire is,
$F = \mu mg$
The formula for the centripetal force of the cycle is given by,
$F = \dfrac{{m{v^2}}}{r}$
Equating the equation (1) and equation (2), we get,
$\mu mg = \dfrac{{m{v^2}}}{r}$
Cancel the mass on both sides, we obtain
$\mu g = \dfrac{{{v^2}}}{r}$

Rearranging the equation for v, then the above equation can be written as,
${v^2} = \mu gr$
Taking square root on both sides, then
$v = \sqrt {\mu gr}$
Substituting the coefficient of friction $\mu = 0.25$, acceleration due to gravity $g = 10m{s^{ - 2}}$, and radius $r = 40m$ in the above equation, we get,
$v = \sqrt {0.25 \times 10 \times 40}$
On multiplying the above equation, then
$v = 10\,m{s^{ - 1}}$
Therefore, the maximum speed is $10\,m{s^{ - 1}}$.

Hence, option D is the correct answer.

Note:Here in this problem, there are two forces acting, that is, one is centripetal force because the cycle moves in the circular path and the other force is the frictional force. According to the law of conservation of energy, when the two forces acting on the system are said to be equal.