Calculating Plane's Work and Force on Aircraft Carrier

[ajc9387]

A plane weighing 220 kN (25 tons) lands on an aircraft carrier. The plane is moving horizontally at 63 m/s (141 mi/h) when its tailhook grabs hold of the arresting cables. The cables bring the plane to a stop in a distance of 89 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?


This question is confusing me. I was under the impression that force needed to be calculated in order to find the work, but clearly they want me to find the work first. So, i went about trying to solve the problem:

W= Force times distance

The distance is equal to 89 meters. The force and the work, though, are both questions to be answered. Force = m times a...mass is 25000 kg, but acceleration isn't constant. Wherever I go with this problem i hit a dead end. Someone please help.

 

[Doc Al]

While multiplying force by distance is the most direct way of calculating work, it won't help you here since you don't know the force. (As you realize.) In fact, part b asks you to find the force.

But there are other ways to deduce the work done and thus the force of the cables. Hint: Consider the Work-Kinetic Energy theorem. (Look it up if you have to.)

 

[kyle8921]

I can provide a response to this content by breaking down the problem and providing a step-by-step solution.

Firstly, we need to determine the initial kinetic energy of the plane as it lands on the aircraft carrier. This can be calculated using the formula:

Kinetic energy = 1/2 * mass * velocity^2

Therefore, the kinetic energy of the plane is:

KE = 1/2 * 220,000 N * (63 m/s)^2 = 415,260,000 joules

(a) To calculate the work done on the plane by the arresting cables, we need to use the formula:

Work = force * distance

Since the distance is given as 89 m, we just need to find the force exerted by the cables.

(b) The force exerted by the cables can be found using Newton's Second Law, which states that force is equal to mass times acceleration (F=ma). In this case, the mass of the plane remains constant at 220,000 N, but the acceleration changes as the plane comes to a stop.

To find the acceleration, we can use the formula:

Final velocity^2 = Initial velocity^2 + 2*a*d

Where d is the distance traveled and a is the acceleration.

Rearranging the formula, we get:

a = (Final velocity^2 - Initial velocity^2) / (2*d)

Plugging in the values, we get:

a = (0 - (63 m/s)^2) / (2 * 89 m) = -0.2 m/s^2

Since the plane is decelerating, the acceleration is negative.

Now, we can calculate the force exerted by the cables using F=ma:

F = 220,000 N * (-0.2 m/s^2) = -44,000 N

Since the force is negative, it means that the cables are exerting a force in the opposite direction of the plane's motion, which is necessary to bring it to a stop.

Finally, we can calculate the work done on the plane by the arresting cables:

Work = force * distance = (-44,000 N) * (89 m) = -3,916,000 joules

The negative sign indicates that the work is being done against the motion of the plane, as it is being brought to