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 88.0 N/m. If the hose is stretched by 4.20 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

Answer:

The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules

Explanation:

The given spring constant of the of the spring, k = 88.0 N/m

The length by which the hose is stretched, x = 4.20 m

For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose

The elastic potential energy, P.E., of a compressed spring is given as follows;

P.E. = 1/2·k·x²

∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²

1/2 × 88.0 N/m × (4.20 m)² = 776.16 J

The work done on the hose = The potential energy given to hose, P.E. = 776.16 J