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).
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?
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