A 16,000 kg F-18 (jet fighter) lands at 52 m/s on an aircraft carrier, its tail hook snags the cable to slow it down. The cable is attached to a spring with a spring constant of 60,000 N/m. How far does the spring stretch to stop the jet?

To find the distance the spring stretches to stop the jet, we can use the concept of conservation of energy.

1. Determine the initial kinetic energy of the jet before it lands. The formula for kinetic energy is given by KE = 0.5 * mass * velocity^2.
KE = 0.5 * 16000 kg * (52 m/s)^2

2. The energy of the spring is potential energy, given by PE = 0.5 * k * stretch^2, where k is the spring constant and stretch is the distance the spring stretches.
PE = 0.5 * 60000 N/m * stretch^2

3. Since the energy of the jet needs to be dissipated entirely by the spring, we equate the initial kinetic energy to the potential energy of the spring:
0.5 * 16000 kg * (52 m/s)^2 = 0.5 * 60000 N/m * stretch^2

4. Rearrange the equation to solve for stretch:
stretch^2 = (0.5 * 16000 kg * (52 m/s)^2) / (0.5 * 60000 N/m)

5. Calculate the square root of both sides to find the stretch:
stretch = √((0.5 * 16000 kg * (52 m/s)^2) / (0.5 * 60000 N/m))

6. Plug in the values and calculate:
stretch = √((0.5 * 16000 kg * (52 m/s)^2) / (0.5 * 60000 N/m))
stretch ≈ √(166400000 J / 30000 N/m)
stretch ≈ √5546.67 m^2)
stretch ≈ 74.5 m

Therefore, the spring stretches approximately 74.5 meters to stop the jet.