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When working with springs, understanding the spring constant is crucial. It is denoted by the symbol \( k \) and measures how stiff or flexible a spring is. The unit of a spring constant is newtons per meter (N/m). A higher value of \( k \) indicates a stiffer spring that requires more force to compress or extend by a certain amount. Conversely, a lower \( k \) value means the spring is more flexible.
The spring constant is unique to each spring and is influenced by factors like material and coil density. If you're ever trying to determine how much force you need to compress a spring a specific amount, or how much it will move when a force is applied, knowing \( k \) is fundamental.
In the original exercise, the spring has a spring constant of \( 56 \mathrm{~N/m} \). This tells us that for every meter the spring is compressed or extended, \( 56 \mathrm{~N} \) of force is needed.

The relationship between force and displacement for springs is captured by Hooke's Law, expressed by the formula \( F = kx \). In this formula:

• \( F \) is the force applied to the spring in newtons (N).
• \( k \) is the spring constant in newtons per meter (N/m).
• \( x \) is the displacement or the compression/extension of the spring in meters (m).

Hooke's Law shows a direct relationship wherein applying a greater force results in a larger displacement, assuming the spring’s elastic limit is not exceeded.
In practical terms, this means if you know two of these variables, you can solve for the third. For instance, if you know the force and spring constant (like in our exercise), you can calculate how much the spring will be displaced by using the formula rearranged as \( x = \frac{F}{k} \). This is exactly what was done in the workout of the original exercise.

Compression of a spring occurs when a force is applied, causing it to shorten in length. This happens when the force compresses the spring, creating potential energy stored within the spring until the force is released.
In the original exercise, a force of \( 6.1 \mathrm{~N} \) compresses a spring with a spring constant of \( 56 \mathrm{~N/m} \). By applying the relationship outlined in Hooke’s Law, we can calculate the exact distance the spring is compressed:
The formula used is \( x = \frac{F}{k} \), where substituting the given values gives \( x = \frac{6.1}{56} \), resulting in approximately \( 0.109 \) meters. Therefore, the spring is compressed by about \( 0.109 \) meters, which is the same as \( 10.9 \) centimeters.Remember, actual spring behavior can diverge from this ideal once the elastic limit is surpassed, at which point Hooke's Law no longer applies.