- #1

physicsss

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(a) Determine the speed of the mass when the spring returns to its normal length (x = 0).

(b) Determine the speed of the mass when the spring returns to half its original extension (x = xmax/2).

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- Thread starter physicsss
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- #1

physicsss

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(a) Determine the speed of the mass when the spring returns to its normal length (x = 0).

(b) Determine the speed of the mass when the spring returns to half its original extension (x = xmax/2).

- #2

Diane_

Homework Helper

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b) Ditto.

- #3

Pyrrhus

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Conservation of Mechanical Energy

[tex] \Delta K + \Delta \Omega = 0 [/tex]

or

[tex] \Delta E = 0 [/tex]

- #4

physicsss

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what does Omega mean?

- #5

Pyrrhus

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Sorry, i was busy, it means Potential Energy.

- #6

physicsss

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, substituting kx for f i got sqrt(Fx/m). This is right.

But for part b, 1/2mv^2=1/2k*(x/2)^2, and when I carried this all out, I got v=1/2*sqrt(Fx/m)...but I got it wrong.

- #7

Pyrrhus

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You should use the point where it's at its max potential energy (max length) and equal it to its energy at half lenght.

[tex] \frac{1}{2}kx_{max}^{2} = \frac{1}{2}mv^2 + \frac{1}{2}k(\frac{x_{max}}{2})^{2} [/tex]

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