Question

A car is moving along a straight horizontal road with a speed\[{v_o}\]. If the coefficient of friction between the tyres and the road is μ, the shortest distance in which the car can be stopped is?
A. $\dfrac{{{\text{v}}_0^2}}{{\mu {\text{g}}}}$
B. ${\left( {\dfrac{{{{\text{v}}_0}}}{{\mu {\text{g}}}}} \right)^2}$
C. $\dfrac{{{\text{v}}_0^2}}{{3\mu {\text{g}}}}$
D. $\dfrac{{{\text{v}}_0^2}}{{2\mu {\text{g}}}}$

Answer 1

Hint: Friction is the force that prevents solid surfaces, fluid layers, and material components sliding against each other from moving in the same direction. Friction comes in a variety of forms. The force that opposes the relative lateral motion of two solid surfaces in contact is known as dry friction. Static friction ("stiction") between non-moving surfaces and kinetic friction ("friction") between moving surfaces are the two types of dry friction. This is a notion that we utilise in this article.

Complete step by step solution:
The coefficient of friction (COF) is a dimensionless scalar number that defines the ratio of the force of friction between two bodies to the force pushing them together. It is frequently represented by the Greek letter$\mu $.
$f = \mu F$
f = friction force
$\mu $ = Coefficient of friction
F = normal force
The coefficient of friction varies depending on the materials used; ice on steel, for example, has a low coefficient of friction, but rubber on pavement has a high coefficient of friction. Friction coefficients range from near zero to more than one. Friction between metal surfaces is greater between two surfaces of similar metals than between two surfaces of different metals, according to the axiom.
The kinetic energy of the body is equal to the work done against frictional force.
A body of mass m possesses kinetic energy when it travels at a velocity of v as$\dfrac{1}{2}m{v^2}$ .
This energy is used to work against the frictional force that exists between the car's tyres and the road.
kinetic energy is the amount of work done against a friction force.
$\dfrac{1}{2}m{v^2} = \mu {\text{ F}}$
We know that F = ma we get
$\dfrac{1}{2}m{v^2} = \mu {\text{ mgs}}$
where s is the distance in which the car is stopped
suppose $v = {v_o}$
upon simplification we get
\[s = \dfrac{{{v_o}^2}}{{2\mu g}}\]
Hence option D is correct.

Note:
$\mu = {\mu _{\text{s}}}$Where${\mu _{\text{s}}}$ is the coefficient of static friction for surfaces that are at rest relative to each other. This is generally more substantial than the kinetic equivalent. The combined effects of material deformation characteristics and surface roughness, both of which have their origins in the chemical bonding between atoms in each of the bulk materials and between the material surfaces and any adsorbed material, determine the coefficient of static friction exhibited by a pair of contacting surfaces. The amount of static friction is known to be influenced by the fractality of surfaces, a characteristic that describes the scaling behaviour of surface asperities.