Question

A car of mass ‘m’ is driven with acceleration ‘a’ along a straight level road against a constant external resistive force ‘R’. When the velocity of the car is ‘V’, the rate at which the engine of the car is doing the work will be
$\begin{array}{c}\text{A}\text{. }RV\\ \text{B}\text{. }maV\\ \text{C}\text{. }\left(R+ma\right)V\\ \text{D}\text{. }\left(ma-R\right)V\end{array}$

Answer 1

Hint: The resistive force R acts on the car in such a way that it produces deceleration in the car as it is acting in the direction opposite to the direction of motion of the car. In order to overcome this force the car has to do extra work.

Formula Used:
Newton’s second law of motion:
$F=ma$
Power:
$P=F\times v$
 where v is the velocity of motion.

Complete step-by-step answer:
We are given a car of mass m moving with acceleration a. There are two forces acting on the car.
Force due to its mass and motion given by Newton’s second law of motion,
$F=ma$.
The constant external resistive force R.
When an external force is acting on a body, then the resultant effect is that the body needs to do more work, which means that total force adds up and is the sum of force due to Newton’s second law and the external force.
Therefore, for the car the total force acting on it is given as
$$\begin{gathered} F^{\prime }=F+R \\ \Rightarrow F^{\prime }=ma+R \end{gathered}$$
Therefore, the rate at which the engine is doing work, which is called the power of engine, is given as a product of total force acting on the car and the velocity of the car.
$P=F^{\prime }\times v$

Substituting value of F’, we get
$P=(ma+R)v$
Hence, the correct answer is option C.

Note: The resistive force adds up to the total force as the engine needs to do more work to overcome it. If the force is not resistive in nature and is aiding the motion of the car then the total force will reduce and only then answer would be option D.