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What is the spring constant for a spring if pulling it back 0.391 m causes a force in the opposite direction of 0.922 N?0.361 N-0.361 N2.36 N-2.36 N

User NealeU
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1 Answer

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We are asked to determine the spring constant given that it is stretched a distance of 0.391 meters and there is a restoring force of 0.922 Newtons.

To do that we will use Hooke's law:


F=-kx

Where:


\begin{gathered} F=\text{ restoring force} \\ k=\text{ spring constant} \\ x=\text{ deformation} \end{gathered}

Now, we solve for the spring constant by dividing both sides by "-x":


(F)/(-x)=k

now, we plug in the values:


(-0.922N)/(-0.391m)=k

Now, we solve the operations:


2.36N=k

Therefore, the spring constant is 2.36 Newtons.

User Junho Park
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