Posted by sandhu on Sunday, November 28, 2010 at 10:03pm.
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?
Physics - drwls, Monday, November 29, 2010 at 3:08am
(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
Physics - sandhu, Monday, November 29, 2010 at 7:30am
It is a timely help.Thanks
Physics - sandhu, Monday, November 29, 2010 at 8:39am
Should the initial stretch of 5.6 ft be added to X to find the maximum distance the band is stretched?
You have to subtract out the initial energy.
1/2 m v^2= 1/2 k(x^2-5.6^2)
With this, x will be the final total stetch, which includes the original)
This equation reads: the kinetic energy is equal to the change in potential energy stored in the band.
Now a note: Airplanes are NOT stopped with elastic bands, my God, they would be tossed backward off the carrier. The cables that aircraft hook to attached to hydraulic energy absorbent devices (very similar to a piston with a hole in it).
To find the maximum distance the band is stretched, you need to solve for X in the equation (1/2)MV^2 = (1/2)kX^2.
First, you need to convert the weight of the airplane from pounds to slugs. Since force is measured in pounds and distance is measured in feet, the weight in pounds needs to be converted to slugs. The weight of the airplane is given as 7200 pounds, which is equal to 223.6 slugs.
Next, plug in the values into the equation. The mass (M) is 223.6 slugs and the speed (V) is given as 72 ft/s. The elastic constant (k) is given as 998 lb/ft. We need to convert this to slugs/ft by dividing by the acceleration due to gravity which is approximately 32.2 ft/s^2. This gives us a value of approximately 31.0 slugs/ft.
Simplifying the equation, (1/2)(223.6 slugs)(72 ft/s)^2 = (1/2)(31.0 slugs/ft)X^2.
Solving for X, we can rearrange the equation to get X = sqrt((223.6 slugs)(72 ft/s)^2 / (31.0 slugs/ft)).
Evaluate this expression using a calculator to get the value of X.
Now, to answer your question about whether the initial stretch of 5.6 ft should be added to X, the answer is no. The equation you are using assumes the initial stretch is already taken into account and the variable X represents the additional distance the band is stretched beyond its initial stretch. Therefore, you do not need to add the initial stretch to X when finding the maximum distance the band is stretched.