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mybrohshi5
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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 = 1/2mv2
W = 1/2kx2
F = -kx
P = F*v
The Attempt at a Solution
I found the velocity
10 = 1/2(3)(v2)
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