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 do-ms 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 85.0 N/m. If the hose is stretched by 2.90 m and then released, how much work docs the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length?

Explanation / Answer

While the surgical hose is obeying Hooke's law the work done on them while being stretched is all stored as strain energy and all released as work done on the balloon (all transferred to KE)

WD on the hose = av force x displacement = ½ke x e

= ½(85N/m x 2.90m) x 2.90m
=WD = 357.425 J

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.