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A horizontal massless spring with spring constant
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Answer #1

from work -energy theorem,


potential enenrgy of 1st spring + work done by friction = potential energy of 2nd spring


300 x 0.20^2 /2 + ( - 0.2 x 1 x 9.8 x 2 ) = 600x^2 /2


x = 0.0833 m or 8.33 cm

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