A spring has natural length 2 cm. When a mass of 5 g is attached, its length is 3.5 cm. What is the extension of the spring?

Step 1: State Hooke's Law.
Hooke's Law states that the extension of a spring is directly proportional to the applied force, provided the elastic limit is not exceeded. Mathematically, this is expressed as $F = ke$, where $F$ is the force, $k$ is the spring constant, and $e$ is the extension.

b) Step 1: Identify the natural length and extension for the second spring.
From the diagram, we assume the first spring (length $2 \text{ cm}$ with $5 \text{ g}$ mass) represents the natural length ($L_0$) of the spring, so $L_0 = 2 \text{ cm}$. The $5 \text{ g}$ label on the first spring is likely indicating the type of spring or is extraneous if it's the natural length.
For the second spring, a mass of $5 \text{ g}$ is attached, and the total length is $3.5 \text{ cm}$.
The extension ($e$) is the total length minus the natural length:

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