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November 27, 2014

November 27, 2014

Posted by **Kristen** on Sunday, November 18, 2007 at 6:12pm.

At what rate is the spring doing work on the ladle as the ladle passes through its equilibruim position?

And

At what rate is the spring doing work on the ladle when the spring is compressed .1m and the ladle is moving away from the equilibrium position?

- I really dont get this. physics -
**bobpursley**, Sunday, November 18, 2007 at 6:28pmI presume you are in calculus based physics.

Rate of work= d work/dt= d/dt (1/2 kx^2)= kx dx/dt = kx * velocity

But velocity= sqrt(2*KE/mass)

where KE= 10-1/2 kx^2 so

**rate of work= kx*(sqrt2/mass (10-1/2kx^2)**

check my thinking.

- I really dont get this. physics -
**Kristen**, Sunday, November 18, 2007 at 6:31pmok..that helps..thanks

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