Answer to Question #121671 in Quantum Mechanics for Vusimuzi Mohlamme

Question

An airplane flies 120 km at a constant altitude in a direction 30.0° north of east. A wind is blowing that results in a net horizontal force on the plane due to the air of 2.40 kN in a direction 10.0° south of west. How much work is done on the plane by the air?
Note on the meaning of 1E8 in answers: 1E8 is an alternative way to write "10 to the power of 8". So 1E8 is 100 000 000. It is a convenient notation when superscripts are difficult to accommodate.

Answer 1

Let, the distance traveled by airplane $d=120km=12\times 10^4m$ and force acting on it by air is $F=2.40kN=2.4\times10^3N$ ,

Now, draw the free body diagram

As, we know that , work done by a force is

$W=\overrightarrow{F}\cdot\overrightarrow{d}=Fd\cos(\theta)$

Where, $\theta$ angle between $\overrightarrow{F}$ and $\overrightarrow{d}$ .

Here,

$\theta=60^{\circ}+10^{\circ}=70^{\circ}$

Thus,

$W=2.4\times10^3\times12\times10^4\cos(70^{\circ})\\ \implies W=9.85\times10^7J$