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 $6.171 N/m$.