Answer to Question #118112 in Classical Mechanics for Lizwi

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.
1 : -2.71 × 1E8 J
2 : -0.985 × 1E8 J
3 : -221 × 1E8 J
4 : 221 × 1E8 J
5 : 0.821 × 1E8

Answer 1

30° N of E means that the airplane's displacement vector was 30° above E-axis. 10° S of W means that the wind's force vector is 10° belox W-axis, therefore, the angle between the displacement and force vectors is 160°. As we know, the work is given by

$$W=Fd\text{ cos}\theta=\\=2400\cdot120000\cdot\text{cos}160^\circ=-271\text{E}8\text{ J}.$$