Question

A one-ton car moves with a constant velocity of $15m{{s}^{-1}}$ on a rough horizontal road. The total resistance to the motion of the car is 12% of the weight of the car. The power required to keep the car moving with the same constant velocity of $15m{{s}^{-1}}$ is [Take $g=10m{{s}^{-2}}$]
A.) 9 kW
B.) 18 kW
C.) 24 kW
D.) 36 kW

Answer 1

Hint: Power is the amount of work done per unit time. Therefore, we can find the formula from this to find the power of the car to maintain the constant velocity. For the constant velocity, the car has to overcome the resistive force. So, we can equate this and put it in the power equation. Finally, we can assign the given information in the equation.

Formula used:

$\text{Power = Net force on the car }\times \text{ Velocity}$
$F=mg$, where F is the force, m is the mass and g is the acceleration due to gravity.

Complete step by step answer:

According to the question, the car is moving with the velocity of $15m{{s}^{-1}}$. But the car experiences a resistance due to the friction. It is about 12% of the mass of the car. To move with a constant velocity, the car has to overcome the frictional force. So the power required for the movement can be written as the product of net force on the car and the velocity of the car.

$\text{Power = Net force on the car }\times \text{ Velocity}$

$F=mg$, where F is the force, m is the mass and g is the acceleration due to gravity.

Since the car has to overcome the frictional force for the constant velocity, we can treat the frictional force as the net force acting on the car. So that net force on the car can be written as,

$P=\left[ \dfrac{12}{100}\text{ }\times \text{mg} \right]\times v$

The mass of the car is 1 ton and the velocity is $15m{{s}^{-1}}$. We can assign these details to the equation.

$P=\left[ \dfrac{12}{100}\times {{10}^{3}}\times 10 \right]\times 15$
$P=1200\times 15$
$P=18000W=18kW$

Therefore, the correct option is B.

Additional information:
Power is the amount of energy consumed to do work in a unit time. But it can be used in various fields. So it has several types of equations.

$P=\dfrac{E}{t}=\dfrac{W}{t}$, where P is the power, E is the energy, W is the work done and t is the time. In this problem, the work is taken as a product of force and displacement. So the ratio of displacement and time gives the velocity. That’s why we multiplied the force and velocity to find the power of the car.

In electrical circuits, power can be found by using the following formulas.

$P=\dfrac{V}{I}=\dfrac{{{V}^{2}}}{R}={{I}^{2}}R$, where V is the voltage applied, I is the current in the circuit and R is the resistance.

Note: Here we found the formula to find the power is formulated from the ratio of work done per unit time.

$P=\dfrac{W}{t}$

Work done can be written as the product of force and displacement.

$P=\dfrac{F.x}{t}$

As we know, the ratio between displacement and time is known as the velocity. Therefore,

$P=F\times v$

The candidates are advised to learn some big units like a ton, quintal etc. Otherwise, the answer will be wrong because most of the other units will be in the SI system. So, we have to convert all units into SI systems for easy calculations.