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A 1.7 kg object is attached to a horizontal spring of force constant k=3200Nm. The spring...

A 1.7 kg object is attached to a horizontal spring of force constant k=3200Nm. The spring is stretched 53 cm from the equilibrium position and released. What is its maximum speed?

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Answer #1

E = 1/2 kA​​​​​2​​​​ = 1/2 mv​​​​​2 + 1/2 kX​2

V = √((k/m)*( A​​​​​​2 - X​​​​​​2 ))

V = √((3200/1.7)*( 0.532 - 0.2652 )

V = 19.9 m/s

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