- #1

- 319

- 0

(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).

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter physicsss
- Start date

- #1

- 319

- 0

(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

- 390

- 0

b) Ditto.

- #3

Pyrrhus

Homework Helper

- 2,179

- 1

Conservation of Mechanical Energy

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

or

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

- #4

- 319

- 0

what does Omega mean?

- #5

Pyrrhus

Homework Helper

- 2,179

- 1

Sorry, i was busy, it means Potential Energy.

- #6

- 319

- 0

, 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

Homework Helper

- 2,179

- 1

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]

Share:

- Replies
- 1

- Views
- 9K