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## Homework Statement

A 3.00 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 468 N/m) the other end of which is fixed. The ladle has a kinetic energy of 10.00 J as it passes through its equilibrium position (the point at which the spring force is zero).

**At what rate is the spring doing work on the ladle when the spring is compressed by 10.0 cm and the ladle is moving away from the equilibrium position?**

## Homework Equations

K = 1/2mv

^{2}

W = 1/2kx

^{2}

F = -kx

P = F*v

## The Attempt at a Solution

I found the velocity

10 = 1/2(3)(v

^{2})

v = 2.58

I found the force next

F = -468(.1)

F = -46.8

Then i found the power which is the rate of the spring doing work on the ladle

P = -46.8(2.58)

P = -120.74 W

Can anyone tell me where i went wrong here?

Thank you