91Ó°ÊÓ

The spring constant, often denoted as \( k \), is a measure of a spring's stiffness. It tells us how hard or easy it is to stretch or compress a spring. The unit for the spring constant is Newtons per meter (N/m). The bigger the spring constant, the stiffer the spring is. A soft spring will have a small spring constant, while a stiffer spring will have a larger one.
When we apply a force to a spring, the spring constant helps us understand how much the spring will stretch. This relationship is described by Hooke's Law, summarized by the equation \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the distance the spring is stretched or compressed.
In our specific exercise, a force of 5.0 N stretches the spring by 2.0 cm, or 0.02 m. Using Hooke's Law, we find the spring constant to be \( k = \frac{5.0}{0.02} = 250 \, N/m \). This means for every meter we stretch the spring, 250 Newtons of force is needed. For short stretches, it translates to one needing 5 N for every 2 cm.

• Hooke's Law: \( F = kx \)
• Spring constant \( k = 250 \, N/m \)
• Higher \( k \), stiffer spring

Elastic potential energy is the energy stored in an object that can be stretched or compressed, like a spring. When we stretch a spring, work is done on it, and this work gets stored as potential energy. If we release the spring, this energy can do work on other objects.
We calculate elastic potential energy using the formula \( U = \frac{1}{2} k x^2 \). The potential energy depends on both the spring constant \( k \) and the square of the distance \( x \) the spring is stretched. So, more stretch means more energy stored.
In our scenario, when the spring is initially stretched by 2.0 cm (0.02 m), the elastic potential energy is \( U_{initial} = \frac{1}{2} \times 250 \, N/m \times (0.02 \, m)^2 = 0.05 \, J \). Upon extending the spring by an additional 4.0 cm, the total stretch becomes 6.0 cm (0.06 m), giving us a final energy of \( U_{final} = \frac{1}{2} \times 250 \, N/m \times (0.06 \, m)^2 = 0.45 \, J \).
Thus, the energy increases significantly, illustrating how doubling the stretch can lead to a dramatic increase in energy.

• Potential Energy Formula: \( U = \frac{1}{2} k x^2 \)
• Initial Energy: 0.05 J at 2 cm
• Final Energy: 0.45 J at 6 cm
• Energy grows with the square of distance

To find out how much additional force is needed to further stretch a spring, Hooke's Law is again our trustworthy guide. We have already determined a spring constant of 250 N/m from our previous interaction. Now, we need to calculate the force required for an additional stretch of 4 cm which translates the total stretch to 6 cm (or 0.06 m).
By plugging these values back into Hooke's Law \( F = kx \), we determine the force required for this total stretch. Therefore, \( F = 250 \, N/m \times 0.06 \, m = 15 \, N \). This means that for the additional stretch, you need to apply a total force of 15 N.
This result further illustrates how the amount of force required depends on the spring constant and the distance stretched.

• Total Force Required: 15 N
• More stretch requires more force
• Hooke's Law helps predict necessary force