This question aims to find the power dissipated by a drag force when velocity is kept constant.
Drag force is a force experienced by any object moving with a certain velocity. If objects do not experience any kind of force, then they will be moving like a breeze. Drag force quadratically increases with the velocity. At higher velocities, an object needs more force to move forward. A higher volume of gas is dissipated when an object moves with a certain velocity.
Drag force is experienced by fast-moving vehicles like airplanes, trains, cars, etc. The force to move gas molecules increases with the movement of these vehicles. The drag force is represented as:
\[F_d = C_dAv^2\]
In the above formula, $A$ represents the cross-sectional area of the vehicle, $v$ represents the velocity, and $C_d$ is the coefficient of drag. The square of velocity means that drag force increases with a moving object.
Expert Answer
A car is moving with maximum velocity $v_o$, where $v_o$ is limited by drag force which is proportional to the velocity square. The maximum power of this engine is $P_o$. When the engine of this car is modified, then the power will become $P_1$
This new power of the modified engine is now ten times greater than the previous power. It is represented as ($P_1$ = $100$ % $P_o$).
If we assume that the top speed is limited by air drag, then the square of velocity is proportional to the drag force. The percentage at which the top velocity of the car is increased:
Relating power and drag force by:
\[Power = F_d \times v\]
\[P = – F_d v\]
Drag force is acting opposite to the moving car, so $\cos$ $(180°)$ = $-1$.
\[P = – C_d A v^2 /times v\]
\[P = – C_d A v^3\]
The initial power is $P_o$, so its magnitude can be written as:
\[P_o = C_dAv_o^{3}\]
\[P_1 = 110% P_o\]
\[P_1 = \frac{110}{100} P_o\]
In magnitude, $P_1$ is written as:
\[P_1 = C_d A v_1^{3}\]
\[C_d A v_1^{3} = C_d A v_o^{3} \times \frac{110}{100}\]
\[v_1^{3} = \frac{11}{10} \times v_o^{3}\]
\[v_1 \thickapprox 1.0323 v_o\]
\[= \frac{v_1 – v_o}{v_o}\]
\[= \frac{1.0323 v_o – v_o}{v_o}\]
\[= 0.0323\]
Numerical Solution
The increase in percentage is $3.23 \%$.
A percentage increase is $3.2$ % if we consider up to two significant numbers.
Example
Consider a car whose shape shows an aerodynamic drag coefficient that is $C_d$ = $0.33$ and the area of the car is $3.4 m^2$.
If we further assume that drag force is proportional to $v^2$ and we neglect other sources of friction where $v^2$ is $5.5 m/s$
By calculating the the drag force:
\[F_d = C_d A v^2\]
\[F_d = 0.33 \times 3.4 \times 5.5 \]
\[F_d = 6.171 N/m\]
The drag force $F_d$ is $6.171 N/m$.