During spring semester at MIT, residents of the parallel buildings of the East C

Question

During 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 and has a spring constant of 105 N/m. If the hose is stretched by 5.2 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?

Explanation / Answer

This would be a simple work=force*distance problem if the force was constant during acceleration, but the spring force is F= -k * x so the problem needs some calculus.

Picking the origin at the starting position of the balloon, with x as the position, and defining d to be the width of the room, the spring force is F=k*(d-x). The work over any small interval is
dWork = d(force*distance) = k*(d-x)*dx
Integrate to get
Work = k*x - k*x^2/2 evaluated over 0 to d, so
Work = k*d - k*d^2/2
Use d=4.6m and k=104N/m.

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