Question

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.)

Short answer

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 \(605,080.87 \, N\).

Step-by-step solution

(Understanding the Given Information)

(First, let's recall the information given in the problem: The plane's initial weight is \(220 \, kN\), its initial velocity is \(67 \, m/s\), and it comes to a stop over a distance of \(84 \, m\). We'll use these values in our calculations.)

(Calculate the Initial and Final Kinetic Energies)

(We need to find out the initial and final kinetic energies of the plane. Remember, kinetic energy is given by the formula KE = \(\frac{1}{2}mv^2\). The initial kinetic energy (KE_initial) of the plane is given by \(\frac{1}{2}M(v_1)^2\) where \(m\) is the mass of the plane \((\frac{220 \, kN}{9.81 \, m/s^2})\), and \(v_1\) is the initial velocity of the plane, \(67 \, m/s\). Since the plane comes to a stop, its final kinetic energy (KE_final) is \(0.\))

(Applying Work-Energy Theorem)

(Now, we will apply the work-energy theorem which states that the work done on an object is equal to the change in its kinetic energy. Using the formula, W = \(\Delta\)KE = KE_final - KE_initial, we can find the work done by the arresting cables on the plane.)

(Calculating the Work Done)

(Plug in the values of initial and final kinetic energies into the work-energy theorem formula. W = \(0 - \frac{1}{2}M(v_1)^2\). Solve for W to get the work done by the arresting cables.)

(Finding the Force Exerted by the Arresting Cables)

(We can find the constant force exerted on the plane by the cables using the definition of work done: \(W = Fd \cos\theta\), where \(F\) is the force, \(d\) is the stopping distance \(84 \, m\), and \(\theta\) is the angle between the force and displacement (in this case, \(\theta = 180^{\circ}\) because the force acts opposite to the displacement). Since \(W\) is negative and the cosine of \(180^{\circ}\) is \(-1\), we can rearrange this equation as \(F = -\frac{W}{d}\). Plug in the values of \(W\) and \(d\) and solve for the force exerted by the arresting cables.) Now, let's compute the values for the work done by the arresting cables and the force exerted by them using the information and formulas we derived above.