91Ó°ÊÓ

The spring constant, often denoted as \( k \), is a measure of a spring's stiffness. It tells us how much force is needed to stretch or compress a spring by a unit of distance, typically meters. The unit of the spring constant is Newton per meter \( (N/m) \).

Hooke's Law states that the force required to compress or stretch a spring by a distance \( x \) from its normal length is proportional to the displacement. This proportionality is what defines the spring constant in the formula:

• \( F = -kx \)

Here, \( F \) represents the force applied, and \( x \) is the change in spring length from its equilibrium position. A higher \( k \) value means a stiffer spring.

In the exercise, we calculated the spring constant by rearranging Hooke's Law to solve for \( k \) using the parameters of the force exerted by gravity when a mass is suspended from the spring.

The force exerted by a spring is determined by how much the spring is stretched or compressed from its original, unstressed length. According to Hooke's Law, this force can be expressed as:

\( F = -kx \)

• \( F \): The force exerted by the spring (measured in Newtons)
• \( k \): The spring constant (characteristic of the specific spring)
• \( x \): The displacement from the equilibrium position (measured in meters)

The negative sign in the equation indicates that the force exerted by the spring is in the opposite direction of the displacement. If you stretch the spring, it pulls back; if you compress it, it pushes away.

In our exercise, when a 2 kg mass was suspended, the spring stretched by 5 cm translating into a certain amount of force exerted in return by the spring, which is proportional to its spring constant.

Gravitational force is the force exerted by the earth to pull objects towards its center. This force is responsible for the weight of an object and is calculated by the equation:

\( F_{gravity} = mg \)

• \( m \): Mass of the object (measured in kilograms)
• \( g \): Acceleration due to gravity (approximately \( 9.8 \, m/s^2 \) on Earth's surface)

This force acts downward on any mass hung from a spring. In our problem, it causes the spring to stretch. When a mass is attached to the spring, it exerts a gravitational force downward, which is countered by the spring force acting upward.

Thus, the equilibrium for a hanging object occurs when the gravitational force equals the spring's force as described by Hooke's Law. This balance allows us to solve for changes in length or force needed.

The equilibrium position of a spring is its rest position, where no external force is applied, and it's neither compressed nor stretched. When a weight is added, it reaches a new equilibrium position where the forces are balanced.

For a spring hanging vertically with a weight, equilibrium is achieved when the gravitational force \( (mg) \) pulling the weight down is equal to the spring force \( (kx) \) pulling upward. This equilibrium is captured by the equation:

• \( F_{gravity} = F_{spring} \)
• Thus, \( mg = kx \)

By understanding this principle, we can find out how much more a spring will stretch when a heavier object is hung from it, by calculating the new \( x \), or change in spring length, given a new mass.

In our exercise, the original length of 10 cm changed to a new length with both a 2kg and 3kg mass, indicating two separate equilibrium positions.