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 112 N/m. If the hose is stretched by 4.70 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:

1237 J

Explanation:

The work done by the hose on the balloon is the work done by a spring which is

W = 1/2k(x₀² - x₁²) where k = spring constant = 112 N/m, x₀ = 4.70 m and x₁ = 0 m.

Substituting the values of the variables into the equation, we have

W = 1/2k(x₀² - x₁²)

W = 1/2 × 112 N/m((4.70 m)² - (0 m)²)

W = 56 N/m(22.09 m² - 0 m²)

W = 56 N/m(22.09 m²)

W = 1237.04 J

W ≅ 1237 J