4.0 J of work are performed in stretching a spring with a spring constant of 2500 N/m. How much is the spring stretched?

asked Nov 6, 2023 202k views

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$energy = elastic \: potential$

$4 = (1)/(2) kx {}^(2)$

$4 = (1)/(2) (2500)x {}^(2)$

$4 = 1250x {}^(2)$

$x {}^(2) = (4)/(1250) = 0.0032 \: meters$

$x = √(0.0032) = 0.05657 \: meters$

Spbnick answered Nov 13, 2023

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