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 with a spring constant of 76.0 n/m. if the hose is stretched by 2.50 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

Hooke's law states:
F=k*x
Where:
F = force
k = spring constant
x = displacement from initial state

We are given information:
k = 76 N/m
x = 2.5 m

We can calculate force:
F=76*2.5 = 190N

Work is defined as product of force and traveled distance:
W=F*d

traveled distance is equal to distance from relaxed to stretched position:
W=190N*2.5m = 475J