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A spring with spring constant of 27 N/m is
stretched 0.12m from its equilibrium position.
How much work must be done to stretch it
an additional 0.059 m? Answer in units of J.

2006-12-13 14:53:45 · 2 answers · asked by Anonymous in Science & Mathematics Physics

2 answers

W=Fd,

F = kx = 27 N/m x 0.059 m = 1.593 N

W = Fd = 1.593 N x 0.059 m = 0.0940 Nm, 0.0940 J.

2006-12-13 15:04:08 · answer #1 · answered by Alan B 2 · 0 0

Turn to Jesus Christ instead of a physic

2006-12-13 23:03:35 · answer #2 · answered by jay78_er 2 · 0 0

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