To take off, an airplane of mass m = 4 t must reach v = 144…

To take off, an airplane of mass $m = 4\ \mathrm{t}$ must reach $v = 144\ \mathrm{km/h}$ at the end of a runway of length $s = 100\ \mathrm{m}$. The rolling friction coefficient is $\mu = 0.2$. Assuming uniform acceleration, what engine power is needed at take-off?

Explanation of Method

Engine power needed for take-off.

Step 1: Acceleration

\[ v = 40\ \mathrm{m/s}, \]

\[ a = \frac{v^2}{2s} = \frac{1600}{200} = 8\ \mathrm{m/s^2}. \]

Step 2: Thrust

\[ F = ma + \mu mg = 32\,000 + 7848 = 39\,848\ \mathrm{N}. \]

Step 3: Power at take-off

\[ P = Fv = 39\,848 \cdot 40. \]

\[P \approx 1.59 \cdot 10^6\ \mathrm{W} = 1.59\ \mathrm{MW}\]

$F = 39.8\ \mathrm{kN}$ $\mu mg$ $s = 100\ \mathrm{m},\ v = 40\ \mathrm{m/s}$ $P = F v_f = 39\,848 \cdot 40 \approx 1.59\ \mathrm{MW}$

Constant thrust gives the uniform acceleration of 8 m/s². The power $Fv$ grows with the speed and is greatest at lift-off.

Result

\[\boxed{\,P \approx 1.59 \cdot 10^6\ \mathrm{W} = 1.59\ \mathrm{MW}\,}\]

Remark

Check. The original result is correct. It is the power at the moment of take-off, the largest during the run; the average power over the run is half of it, 0.80 MW.