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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 82.0 N/m. If the hose is stretched by 2.70 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?

Respuesta :

Answer:

110.7 J

Explanation:

Hooke's law is represented by the formula:

F = ke  where F is the force in Newton, K is force constant and e is extension in m

work done = 1/2ke² = 1/2 K ( e² - e₀²) and e₀ is the extension at relaxed length

e₀ =0

work done = 0.5 × 82N/m × (2.70 m)² = 110.7 J

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