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

We have given quantities to solve the problem is :-

Mass of airplane with engine M=1.00×105kg

Power P=100MW=100×10^6W

Velocity V=200m/s

Altitude h=12km=12000m

Since this airplane is in motion and it is at a certain height therefore k.e and p.e working on it

Now k.e =(1/2)MV², . p.e=MGH

The total mechanical energy of plane is

M.E=k.e+p.e

M.E=(1/2)mv²+MGH

Now substituting all the required values

Therefore M.E=(1/2)×(1×10⁵)(200)²+(1×10⁵)(9.8)(12000)

M.E=1.37×10¹⁰=1.37×10¹⁰J

Now we have power given . In terms of energy power can be written as P=E/t where E is energy and t is time.so by using this we can calculate the required time

Now 100×10⁶=1.37×10¹⁰/t

t=1.37×10¹⁰/100×10⁶

t=137s

b). Now for time t=950s we can calculate power by using

P=E/t

Where E=1.37×10¹⁰

Therefore P=1.37×10¹⁰/950=14.42MW

Is the Power applied in 950s

C) to calculate F(air) we can calculate Force by engine Fe and displacement first. To calculate force we use formula of Work done by engine

We=Fed

Where Fe is force due to engine and d is displacement

Now D=(1/2)(Vo+Vf)t

Here Vo is 0 therefore Vf=200m/s, t=950s

D=1/2(Vf)t

D=1/2×200×950= 95000=9.5×10⁴m

Now We=Fe×9.5×10⁴

since work done by engine is equal to the energy

Therefore We=1.37×10¹⁰J

Now 1.37×10¹⁰/9.5×10⁴=Fe

Fe=144.21kN

Now the total force acting on plane is by force due to engine Fe and force due to air resistance Far

The total work done is Wnet= Fe+Far is equal to the total energy

Now we use this to calculate for force due to air resistance Far

Wnet=Fe(de)+Far(dar), where de = distance at time 950sand dar= distance at time 1200s

Substituting all value as Wnet=1.37×10¹⁰J, Fe=144.21kN,de=9.5×10⁴ m

dar=1/2×200×1200=12×10⁴m

Now, Far={ (1.37×10^10)-(144.21×9.5×10^4)}÷12×10⁴

Far=1.36×10^10/12×10^4=0.1133×10^3kN