91Ó°ÊÓ

When you think about springs, Hooke's Law comes into play. Hooke's Law explains the relationship between the force applied to a spring and its compression or extension length. In simple terms, it states that the force needed to extend or compress a spring by a certain distance is proportional to that distance. This relationship can be expressed through the formula:
\( F = kx \), where:

• \( F \) is the force you apply in newtons (N),
• \( k \) is the spring constant, representing the stiffness of the spring,
• \( x \) is the displacement from the spring’s natural length (in meters).

Whenever you compress or stretch a spring, the amount of force you need depends on how stiff it is (spring constant) and how far you want to move it (displacement). This principle underlies everyday activities like the exercise problem you're studying.

The spring constant \( k \) tells us how stiff a spring is. A higher spring constant means a stiffer spring. This value is crucial because it helps us determine how much force is required for a certain amount of compression or extension. To find the spring constant, you can use an experiment like in your exercise.
In the given exercise, when the work done is 80.0 J for a compression of 0.200 m, the formula becomes:
\[ W = \frac{1}{2}kx^2 \], which rearranges to:
\[ k = \frac{2W}{x^2} \]

• Here, \( W = 80.0 \) J,
• and \( x = 0.200 \) m,
• leading to \( k \) being computed in N/m.

This formula allows us to understand how much force each movement requires, based on how much energy is stored in the springs.

Understanding how work and energy relate helps in solving problems involving motion and force, like your leg press exercise. The work done on an object is the energy transferred to it by a force moving it over a distance. In physics, the formula for work is
\( W = Fd \), where:

• \( W \) is the work done (in joules),
• \( F \) is the force (in newtons),
• \( d \) is the distance the force moves the object (in meters).

In the context of compressed springs, the work expression becomes more specialized as
\( W = \frac{1}{2}kx^2 \), highlighting how energy is stored in a spring as potential energy. This energy comes into play when calculating how much effort it takes to hold or move the spring further.
Further movement requires additional energy or work, as elucidated in the given exercise: moving the spring another 0.200 m means new calculations for both the total and additional work done, ensuring a deeper understanding of energy dynamics.