A fighter aircraft of mass $m = 10{,}000\text{ kg}$ is flying at a true airspeed $V = 250\text{ m/s}$ in level flight. At this flight condition, the available thrust is $T = 65\text{ kN}$ and the aerodynamic drag is $D = 25\text{ kN}$. Taking $g = 9.81\text{ m/s}^2$, the specific excess power $P_s$ (in $\text{m/s}$) of the aircraft is closest to:

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### 1. Strategy & Theoretical Principles Specific Excess Power ($P_s$) is the excess power per unit weight of the aircraft: $$P_s = \frac{(T - D) V}{W} = \frac{dh}{dt} + \frac{V}{g} \frac{dV}{dt}$$ It represents the maximum potential rate of climb in unaccelerated flight, or the acceleration capability in level flight, or a combination of both along the energy-height curve. ### 2. Analytical Formulation & Calculation Given: - Mass $m = 10{,}000\text{ kg} \implies W = m g = 10{,}000 \times 9.81 = 98{,}100\text{ N}$ - Airspeed $V = 250\text{ m/s}$ - Thrust $T = 65\text{ kN} = 65{,}000\text{ N}$ - Drag $D = 25\text{ kN} = 25{,}000\text{ N}$ Excess thrust: $$T - D = 65{,}000 - 25{,}000 = 40{,}000\text{ N}$$ Excess power: $$(T - D) V = 40{,}000 \times 250 = 10{,}000{,}000\text{ W} = 10\text{ MW}$$ ### 3. Step-by-Step Calculation $$P_s = \frac{(T - D) V}{W} = \frac{10{,}000{,}000}{98{,}100} \approx 101.937\text{ m/s} \approx 101.9\text{ m/s}$$ ### 4. Distractor Forensics - Dividing by mass $m$ instead of weight $W = mg$ gives $1000\text{ W/kg} = 1000\text{ m}^2/\text{s}^3$, forgetting factor of $g$. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Specific excess power $P_s$ has dimensions of velocity ($[P_s] = \text{m/s}$). Always divide excess power by weight $W$, not mass $m$. - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, Chapter 6 (Energy State Approximation).

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