Question

a. How long (in s) would it take a 1.50 ✕ 105 kg airplane with engines that produce of 100 MW of power to reach a speed of 300 m/s and an altitude of 12.0 km if air resistance were negligible? b. If it actually takes 850 s, what is the power applied, in megawatts? c. Given this power, what is the average force (in kN) of air resistance if the airplane takes 1200 s? (Hint: You must find the distance the plane travels in 1200 s assuming constant acceleration.)

Answer 1

Mass of the airplane, M = 1.50 * 10⁵ kg

Power of the engine, P = 100 MW = 100 * 10⁶ W

Speed reached, v = 300 m/s

Altitude, h = 12 km = 12 * 10³ m

Now,

Initially plane was at rest and on the ground,

So,

Total change in energy, E = 1/2*M*v² + M*g*h = (0.5*1.50 * 10⁵*300*300) + (1.50 * 10⁵*9.8*12*10³)

=> E = 67500 * 10⁵ + 176.4 * 10⁸ = 6750 * 10⁶ + 17640 * 10⁶

         = 24390 *10⁶ J

Now,

time taken, t = Total energy / Power = (24390 *10⁶) / (100 * 10⁶)

                      = 243.9 s

(b)

Given,

time taken, t = 850 s

Total energy, E = 24390 *10⁶ J

Thus,

Power required, P = E/t = (24390 *10⁶) / 850

                             = 28.7 *10⁶ W = 28.7 MW

(c)

Time taken, t = 1200 s

Power of the engine, P = 28.7 MW

Thus,

Total energy generated by the engine, E' = P*t = 28.7 * 1200

                                                                   = 34440 MJ

Now,

total energy required, E = 24390 MJ

Thus,

Energy lost, E'' = Energy generated - Energy required

                         = 34440 - 24390

                         = 10050 MJ

Now,

Let the average force of air resistance be F

displacement, d = 12 km = 12 * 10³ m

Now, we know

=> E''= F*d

=> 10050 * 10⁶ = F * 12 * 10³

=> F = (10050 * 10⁶) / (12 * 10³)

         = 837.5 * 10³ N

         = 837.5 kN