Question

(a) How long (in s) would it take a 1.00 ✕ 10⁵ kg airplane with engines that produce of 100 MW of power to reach a speed of 200 m/s and an altitude of 12.0 km if air resistance were negligible?

_? s

(b) If it actually takes 950 s, what is the power applied, in megawatts?

_? MW

(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.)

_? kN

Answer 1

(a) Solve this problem using conservation of energy -

Means -

Etot = KE + PE

P*t = ½m*v² + m*g*h

100 x 10^6 x t =(1/2) x 1 x 10^5 x 200^2 + 1 x 10^5 x 9.8 x 12000

=> 100 x 10^6 x t = 2.0 x 10^9 + 1176 x 10^6

=> 100 x t = 2.0 x 10^3 + 1176

=> t = (2000 + 1176) / 100 = 31.76 s

(b) Suppose, P Watt is the actual power applied.

So, from above -

P x t = 2.0 x 10^9 + 1176 x 10^6

=> P x 950 = (2000 + 1176) x 10^6

=> P = (3176/950) x 10^6 W = 3.34 x 10^6 W = 3.34 MW

(c) Now, work done by the power (3.34 MW) in 1200 s -

W1 = 3.34 x 10^6 x 1200 J

And work done in 950 s is -

W2 = 3.34 x 10^6 x 950 J

So, work done by the air resistance = W1 - W2 = 3.34 x 10^6 x 1200 J - 3.34 x 10^6 x 950 J

= 3.34 x 10^6 x 250 J

Atitide of the plane, h = 12.0 km = 12000 m

Suppose average force of air resistance = F

So, F*h =  3.34 x 10^6 x 250 J

=> F =  (3.34 x 10^6 x 250) / 12000 = 69.58 x 10^3 N = 69.58 kN (Answer)