
The coefficient of kinetic friction between the road and the tyres is $0.2$. Then the shortest distance in which a car be stopped if it was travelling with a velocity of $72{\text{ kmph}}$ is $(g = 9.8{\text{ m}}{{\text{s}}^{ - 2}})$
A. $100{\text{ m}}$
B. $110{\text{ m}}$
C. $105{\text{ m}}$
D. $102{\text{ m}}$
Answer
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Hint: We are given the coefficient of the friction of the road and the tyres and we need to find the shortest distance in which a car can be stopped due to the frictional force. We need to find the acceleration of the vehicle first and find the distance using equations of motion.
Complete step by step answer:
Given, $\mu = 0.2$, $g = 9.8{\text{ m}}{{\text{s}}^{ - 2}}$, $u = 72{\text{ kmph = }}\dfrac{{72 \times 1000}}{{60 \times 60}} = 20{\text{ m}}{{\text{s}}^{ - 1}}$, $v = 0$. Since the car needs to be stopped as required by the question. Now we know that the equation for frictional force must be equal to the force of the vehicle due to its mass and acceleration to stop the vehicle. Therefore we can write
$ \Rightarrow {f_c} = - \mu mg = ma$
$ \Rightarrow a = - \mu g$
Substituting the value of the coefficient of friction we get
$ \Rightarrow a = - 0.2 \times 9.8 = - 1.96{\text{ m}}{{\text{s}}^{ - 2}}$
Now using the third equation of motion we get
$2as = {v^2} - {u^2}$
Where $s$ is the shortest distance to stop the car
Substituting the values we get
$ - 2 \times 1.96 \times s = 0 - 20 \times 20$
$s = \dfrac{{20 \times 20}}{{2 \times 1.96}} \\
\therefore s = 102{\text{ m}}$
Hence option D is the correct answer.
Additional information: Frictional Force refers to the force generated by two surfaces that contact and slide against each other. These forces are mainly affected by the surface texture and quantity of force holding them together in one frame. The angle and position of the object affect the volume of frictional force. This force occurs due to the forces of attraction between the region of contact of the surfaces which has small irregularities.
Note: The negative sign in the frictional force indicates that the frictional force acts in the opposite direction as that of the force due to the acceleration or the direction of the vehicle. The force due to acceleration of the body is given by Newton’s second law of motion.
Complete step by step answer:
Given, $\mu = 0.2$, $g = 9.8{\text{ m}}{{\text{s}}^{ - 2}}$, $u = 72{\text{ kmph = }}\dfrac{{72 \times 1000}}{{60 \times 60}} = 20{\text{ m}}{{\text{s}}^{ - 1}}$, $v = 0$. Since the car needs to be stopped as required by the question. Now we know that the equation for frictional force must be equal to the force of the vehicle due to its mass and acceleration to stop the vehicle. Therefore we can write
$ \Rightarrow {f_c} = - \mu mg = ma$
$ \Rightarrow a = - \mu g$
Substituting the value of the coefficient of friction we get
$ \Rightarrow a = - 0.2 \times 9.8 = - 1.96{\text{ m}}{{\text{s}}^{ - 2}}$
Now using the third equation of motion we get
$2as = {v^2} - {u^2}$
Where $s$ is the shortest distance to stop the car
Substituting the values we get
$ - 2 \times 1.96 \times s = 0 - 20 \times 20$
$s = \dfrac{{20 \times 20}}{{2 \times 1.96}} \\
\therefore s = 102{\text{ m}}$
Hence option D is the correct answer.
Additional information: Frictional Force refers to the force generated by two surfaces that contact and slide against each other. These forces are mainly affected by the surface texture and quantity of force holding them together in one frame. The angle and position of the object affect the volume of frictional force. This force occurs due to the forces of attraction between the region of contact of the surfaces which has small irregularities.
Note: The negative sign in the frictional force indicates that the frictional force acts in the opposite direction as that of the force due to the acceleration or the direction of the vehicle. The force due to acceleration of the body is given by Newton’s second law of motion.
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