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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 92.0 N/m. If the hose is stretched by 3.20 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?

Respuesta :

skyluke89

Answer:

294.4 N

Explanation:

Since the hose obeys Hooke's law, the force it exerts on the balloon in the pouch is given by:

[tex]F=kx[/tex]

where

k is the spring constant

x is the stretching

In this problem,

k = 92.0 N/m

x = 3.20 m

Therefore, the force exerted is

[tex]F=(92.0 N/m)(3.20 m)=294.4 N[/tex]

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