91Ó°ÊÓ

The "force constant," also known as the "spring constant," is a key parameter defining a spring's stiffness. According to Hooke's Law, the force required to compress or stretch a spring is directly proportional to the displacement of the spring. This relationship is expressed by the equation \( F = kx \). Here, \( F \) stands for the force applied, \( x \) is the displacement from the spring's equilibrium position, and \( k \) is the spring constant.

The force constant \( k \) has units of N/m (newtons per meter), which illustrates how much force is needed to elongate or compress the spring by one meter. In our exercise, a force of 800.0 N stretches the spring by 0.200 m.

By rearranging Hooke's Law, we find \( k = \frac{F}{x} \). Substituting in the given values:

• \( F = 800.0 \text{ N} \)
• \( x = 0.200 \text{ m} \)

This gives us \( k = 4000.0 \text{ N/m} \). A high force constant implies a stiffer spring, requiring more force to achieve the same displacement.

When a spring is either stretched or compressed, it stores energy called elastic potential energy. The amount of this energy can be calculated using the formula \( U = \frac{1}{2} k x^2 \). In this formula, \( U \) represents the elastic potential energy, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.

Elastic potential energy increases with greater displacements and stiffer springs (higher \( k \)). In our scenario, the spring constant \( k \) was previously calculated as 4000.0 N/m. To find the potential energy for a displacement of 0.300 m, we substitute:

• \( U = \frac{1}{2} \times 4000.0 \text{ N/m} \times (0.300 \text{ m})^2 \)
• Results in \( U = 180.0 \text{ J} \) both when stretched or compressed by 0.300 m

This indicates that the energy stored depends solely on the magnitude of displacement, not the direction of the displacement.

The work done on a spring when stretching or compressing it refers to the energy transferred into the spring to change its length. This work is equal to the elastic potential energy stored in the spring for a given displacement. Using the formula \( W = \frac{1}{2} k x^2 \), where \( W \) represents the work done, we can determine the energy needed for a specific displacement.

In our exercise, calculating the work done for a 0.200 m stretch involves substituting the known values into the equation:

• \( W = \frac{1}{2} \times 4000.0 \text{ N/m} \times (0.200 \text{ m})^2 \)
• This yields \( W = 80.0 \text{ J} \).

This computation shows how the integral of force over the distance traveled results in the amount of work done, transforming it into stored elastic potential energy.