A plane weighing 236 kN (24.1 tons) lands on an aircraft carrier. The plane is moving horizontally at 66.8 m/s (149.4 mi/h) when its tailhook grabs hold of the arresting cables. The cables bring the plane to a stop in a distance of 84.0 m.
How much work is done on the plane by the arresting cables?
The work done on the plane by the arresting cables is equal to the change in kinetic energy of the plane.
First, we need to calculate the initial and final kinetic energies of the plane:
Initial kinetic energy = 1/2 * m * v^2
= 1/2 * (236000 N / 9.81 m/s^2) * (66.8 m/s)^2
= 444933840 J
Final kinetic energy = 1/2 * m * v^2
= 1/2 * (236000 N / 9.81 m/s^2) * 0^2
= 0 J
The change in kinetic energy is therefore:
Delta KE = final KE - initial KE
= 0 J - 444933840 J
= -444933840 J
Since the change in kinetic energy is negative, this means that work was done on the plane by the arresting cables. The magnitude of the work done is equal to the absolute value of the change in kinetic energy:
Absolute value of Delta KE = 444933840 J
Therefore, the work done on the plane by the arresting cables is 444933840 J.