A spring with a spring constant of 25.0N/m is stretched 4.50m. What is the force that the spring would apply?

asked Oct 16, 2022 118k views

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Step-by-step explanation:

$from \: hookes \: law : \\ force = k \bold{.}e \\ = 25.0 * 4.50 \\ = 112.5 \: N$

CAdaker answered Oct 20, 2022

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