A plane weighing \(220 \mathrm{kN} \quad(25 \text { tons })\) lands on an aircraft carrier. The plane is moving horizontally at $67 \mathrm{m} / \mathrm{s}(150 \mathrm{mi} / \mathrm{h})$ when its tailhook grabs hold of the arresting cables. The cables bring the plane to a stop in a distance of $84 \mathrm{m}$ (a) How much work is done on the plane by the arresting cables? (b) What is the force (assumed constant) exerted on the plane by the cables? (Both answers will be underestimates, since the plane lands with the engines full throttle forward; in case the tailhook fails to grab hold of the cables, the pilot must be ready for immediate takeoff.)

Question: Calculate the work done by the arresting cables and the constant force exerted on the airplane. Solution: 1. Convert the plane's weight to mass: \(m = \frac{220 \, kN}{9.81 \, m/s^2} = 22416.31 \, kg\) 2. Calculate the initial kinetic energy (KE_initial): KE_initial = \(\frac{1}{2}mv^2 = \frac{1}{2} \times 22416.31 \, kg \times (67 \, m/s)^2 = 50,786,772.79 \, J\) 3. Apply the work-energy theorem: W = KE_final - KE_initial = \(0 - 50,786,772.79 \, J = -50,786,772.79 \, J\) 4. Calculate the force exerted by the arresting cables: F = \(-\frac{W}{d} = -\frac{-50,786,772.79 \, J}{84 \, m} = 605,080.87 \, N\) Answer: The work done by the arresting cables is approximately \(-50,786,772.79 \, J\) and the constant force exerted on the airplane by the cables is approximately \(605,080.87 \, N\).