A plane weighing 220 kN (25 tons) lands on an aircraft carrier. The plane is moving horizontally at 67 m/s (150 mi/h) when its tail hook grabs hold of the arresting cables. The cables bring the plane to a stop in a distance of 84 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 engine full throttle forward; in case the tail hook fails to grab hold of the cables, the pilot must be ready for immediate takeoff.)

Plane's mass M = (220*10^3 N)/g

= 22,450 kg

Initial kinetic energy = (1/2) M V^2
= 5.04*10^7 J

(a) Work done by cables = -(initial KE)
(b) Divide work by distance the plane moves while decelerating. That will be the average force exerted

(a) Well, the arresting cables do quite a bit of work, but they are not exactly getting a promotion to "Cable of the Year." To find the work done, we can use the equation W = Fd, where W is the work, F is the force, and d is the distance.

The distance here is 84 m, but we need to convert the force to Newtons. Since 1 kilonewton (kN) = 1000 N, we find that the force is 220,000 N. Now, we can plug in the values into our equation:

W = (220,000 N) * (84 m)
W = 18,480,000 Nm
W = 18,480,000 Joules

So, the arresting cables do approximately 18,480,000 Joules of work on the plane. That's quite a workout!

(b) To find the force exerted on the plane by the cables, we can use the equation F = ma, where F is the force, m is the mass, and a is the acceleration. Since the plane is brought to a stop, its final velocity is 0 m/s.

First, let's convert the mass from tons to kilograms. We know that 1 ton is equal to 1000 kg, so the mass of the plane is 25 * 1000 kg = 25,000 kg.

Next, we can use the equation v^2 = u^2 + 2ad, where v is the final velocity, u is the initial velocity, a is the acceleration, and d is the distance. Rearranging the equation to solve for acceleration, we get a = (v^2 - u^2) / (2d).

Plugging in the values, we have:
a = (0^2 - (67 m/s)^2) / (2 * 84 m)
a = (-4489 m^2/s^2) / (168 m)
a ≈ -26.7 m/s^2 (approximately)

Now, we can find the force:
F = (25,000 kg) * (-26.7 m/s^2)
F ≈ -667,500 N

So, the force exerted on the plane by the cables is approximately -667,500 N. Hmm, negative force? Seems like those cables need a lesson in being more positive!

To calculate the work done on the plane by the arresting cables, we can use the work-energy principle. The work done on an object is equal to the change in its kinetic energy.

(a) First, let's calculate the initial kinetic energy of the plane.
The kinetic energy (KE) is given by the formula: KE = 1/2 * mass * velocity²

The mass of the plane is given as 25 tons, which is equivalent to 25000 kg.
Converting the velocity from m/s to km/h: 67 m/s * 3.6 = 241.2 km/h.

Therefore, the initial kinetic energy of the plane is:
KE_initial = 0.5 * 25000 kg * (241.2 km/h)^2

Next, let's calculate the final kinetic energy of the plane.
Since the plane comes to a stop, the final kinetic energy (KE_final) is zero.

The work done on the plane is equal to the change in kinetic energy:
Work_done = KE_final - KE_initial
Work_done = 0 - KE_initial

(b) To find the force exerted on the plane by the cables, we can use the formula for work:
Work_done = force * distance

We already calculated the work done on the plane, so we can rearrange the formula to find the force:
force = Work_done / distance

Now that we have the values, we can substitute them into the formulas and calculate the answers.

To find the work done on the plane by the arresting cables, we need to calculate the change in kinetic energy of the plane. Work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.

(a) The work done on the plane by the arresting cables is equal to the change in kinetic energy:

Work = Change in Kinetic Energy

The initial kinetic energy (K1) of the plane can be calculated using the mass (m) and its velocity (v1):

K1 = (1/2) * m * v1^2

Given that the mass of the plane is 25 tons, we need to convert it to kilograms (1 ton = 1000 kg):

m = 25 tons = 25 * 1000 kg = 25000 kg

The velocity (v1) of the plane is 67 m/s.

K1 = (1/2) * 25000 kg * (67 m/s)^2

Next, we need to find the final kinetic energy (K2) of the plane, which is zero since the plane comes to a stop:

K2 = 0

The change in kinetic energy is then:

Change in Kinetic Energy = K2 - K1 = 0 - [(1/2) * 25000 kg * (67 m/s)^2]

Therefore, the work done on the plane by the arresting cables is equal to the change in kinetic energy:

Work = Change in Kinetic Energy = - [(1/2) * 25000 kg * (67 m/s)^2]

(b) The force exerted on the plane by the cables can be determined by using the definition of work:

Work = Force * Distance

We know the distance (d) covered by the cables is 84 m and we have already calculated the work done on the plane:

Work = - [(1/2) * 25000 kg * (67 m/s)^2]

Substituting the values into the equation:

- [(1/2) * 25000 kg * (67 m/s)^2] = Force * 84 m

Solving for the force (F):

Force = - [(1/2) * 25000 kg * (67 m/s)^2] / 84 m

Calculate the appropriate values to find the numerical solution.

Please note that this estimate assumes constant force exerted by the cables and does not account for other factors such as air resistance and variations in force during the stopping process.