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uring spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law with a spring constant of 130 N/m. If the hose is stretched by 5.50 m and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length?

Respuesta :

skyluke89

Answer:

1966 J

Explanation:

The work done by the hose on the balloon is equal to the elastic potential energy stored in it:

[tex]W=U=\frac{1}{2}kx^2[/tex]

where

k = 130 N/m is the spring constant

x = 5.50 m is the stretching of the hose before it is being released

If we substitute these numbers into the equation, we find:

[tex]W=U=\frac{1}{2}(130 N/m)(5.50 m)^2=1966 J[/tex]

So, the work done is 1966 J.

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