Question

A 1500 kg car skids to a halt on a wet road where Mk = 0.54.

How fast was the car traveling if it leaves 60-m-long skid marks?

Answer 1

Concepts and reason

The main concept used to solve the problem is work energy theorem.

Initially, the frictional force is equal to the net force causing retardation. Later, equate the two force equations, acceleration can be found. Finally, Use this value the initial velocity is calculated using third law of motion.

Fundamentals

A coefficient of friction is a value that shows the relationship between the force of friction between two objects and the normal reaction between the objects that are involved.

It is defined as the ratio of friction to the normal force.

$\mu = \frac{F}{N}$

Here $\mu$ is coefficient of friction and $N$ is the normal force

According to Newton’s second law of motion, the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object, it is given by,

$F = ma$

Here $F$ is the applied force, $m$ is the mass of object and $a$ is the resulting acceleration.

The equation of velocity distance is given by,

${v^2} - {u^2} = 2as$

Here $v$ is final velocity, $u$ is initial velocity, $s$ is the displacement, and $a$ is the acceleration.

The normal reaction is equal to the weight of the car.

$N = mg$

Here, m is the mass, g is the gravity, and N is the normal force.

The friction force is given by,

$F = \mu N$

Here, F is the force of friction, $\mu$ is the coefficient of friction, and N is the normal force.

Substitute $mg$ for $N$ in expression $F = \mu N$ .

$F = \mu mg$

Equate it to Newton’s second law equation,

$\begin{array}{c}\\\mu mg = - ma\\\\a = - \mu g\\\end{array}$

Negative sign is due to reason that the both forces are equal and opposite. Since, vehicle is retarding then the acceleration will be negative.

Substitute $0.54$ for $\mu$ and $9.81\,{\rm{m}} \cdot {{\rm{s}}^{ - 2}}$ for $g$ in above and simplify

$\begin{array}{c}\\a = - \left( {0.54} \right)\left( {9.81\,{\rm{m}} \cdot {{\rm{s}}^{ - 2}}\,} \right)\\\\ = - 5.2974\,{\rm{m}} \cdot {{\rm{s}}^{ - 2}}\\\end{array}$

Substitute $- 5.2974\,{\rm{m}} \cdot {{\rm{s}}^{ - 2}}$ for $a$ , $0\,{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for $v$ and $60\,{\rm{m}}$ for $s$ in ${v^2} = {u^2} + 2as$ ,

$\begin{array}{c}\\0 = {u^2} + 2\left( { - 5.2974\,{\rm{m/}}{{\rm{s}}^2}} \right)\left( {60\,{\rm{m}}} \right)\\\\{u^2} = 635.688\,{{\rm{m}}^2} \cdot {{\rm{s}}^{ - 2}}\\\\u = 25.212\,{\rm{m}} \cdot {{\rm{s}}^{ - 1}}\\\end{array}$

Ans:

The vehicle was travelling with speed of $25.2\,{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ .