Question

A Boeing 747-8i jet needs 1680 m of runway to land. If its loaded mass is 3.33 × 10^5 kg and it lands at a speed of 275 km/hr, what frictional force is needed to stop it?

Answer 1

We can find the frictional force that is needed to stop the Boeing 747-8i jet, from the work-kinetic energy theorem:

$$\Delta KE=KE_f-KE_i=W=-F_{fr}d,$$ $$0-\dfrac{1}{2}mv^2=-F_{fr}d,$$ $$F_{fr}=-\dfrac{0-\dfrac{1}{2}mv^2}{d},$$ $$F_{fr}=-\dfrac{0-\dfrac{1}{2}\cdot3.33\cdot10^5\ kg\cdot(76.38\ \dfrac{m}{s})^2}{1680}=5.78\cdot10^5\ N.$$