A 7200 lb airplane lands on an aircraft carrier with a speed of 72 ft/s. The plane is caught by an elastic band (k=998 lb/ft) that has an initial stretch of 5.6 feet. What is the maximum distance the band is stretched?
(1/2)MV^2 = (1/2)k X^2
Solve for X
The mass M of the airplane must be in slugs when using this formula, since ft and lb are used for distance and force. 7200 lb weight = 223.6 slugs
It is a timely help.Thanks
Should the initial stretch of 5.6 ft be added to X to find the maximum distance the band is stretched?
To find the maximum distance the band is stretched, we can use the concept of potential energy stored in the elastic band.
The potential energy stored in the elastic band is given by the formula:
Potential energy = 1/2 * k * x^2
Where:
k is the spring constant of the elastic band (998 lb/ft)
x is the distance the band is stretched
Now, let's find the force applied by the airplane on the band using Newton's second law:
Force = mass * acceleration
The weight of the airplane is given as 7200 lb, and we can calculate the acceleration using the formula:
acceleration = Change in velocity / Time
Since the airplane comes to rest when it lands on the aircraft carrier, the change in velocity is equal to the initial speed (72 ft/s) since the final speed is 0. And let's assume it takes 1 second for the airplane to come to rest.
acceleration = 72 ft/s / 1 s
acceleration = 72 ft/s^2
Now, we can calculate the force applied by the airplane:
Force = 7200 lb * 72 ft/s^2
Force = 518,400 lb*ft/s^2
Next, we can equate the force applied by the airplane to the force exerted by the elastic band:
Force = k * x
518,400 lb*ft/s^2 = 998 lb/ft * x
Let's solve for x:
x = 518,400 lb*ft/s^2 / 998 lb/ft
x ≈ 519.04 ft
So, the maximum distance the band is stretched is approximately 519.04 feet.