Question

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?

Answer 1

1966 J

Step-by-step explanation:

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

`W=U=(1)/(2)kx^2`

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:

`W=U=(1)/(2)(130 N/m)(5.50 m)^2=1966 J`

So, the work done is 1966 J.