Thanks in advance for any help someone may provide. I think I\'ve figured out th

Question

Thanks in advance for any help someone may provide. I think I've figured out the answer to this question but it's very important to make sure its correct.

"A certain spring is found NOT to conform to Hooke's Law. The force (in newtons) it exerts when stretched a distance x (in meters) is found to have magnitude 52.8x + 38.4x^2 in the direction opposing the stretch.
(a) With one end of the spring fixed, a particle of mass 2.17 kg is attached to the other end of the spring when it is extended by an amount x = 1.00m. If the particle is then released from rest, compute its speed at the instant the spring has returned to the configuration in which the extension is x = 0.500m."

Thanks for your time.

Explanation / Answer

SEi = SEf 0.5mvi2 + mgh + 0.5kxi2 = 0.5mvf2 + mgh + 0.5kxf2 there will never be any potential energy since its not at a height so those terms go out. it isnt in motion at the beginning so v initial = 0 solve for v final

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