Question

One end of a vertical spring is fastened to the ceiling. A weight is attached to the other end and slowly lowered to its equilibrium position. Show that the loss of gravitational potential energy of the weight equals one-half the gain in spring potential energy. (Why are these two quantities not equal?)

Short answer

The loss of gravitational potential energy of the weight equals one-half of the gain in spring potential energy because the spring doesn't stretch uniformly. The spring tension gradually increases from zero, therefore only half of the gravitational potential energy is converted into spring potential energy.

Step-by-step solution

Express loss of gravitational potential energy

The loss in gravitational potential energy can be calculated using the formula $E_{gp}=mgh$, where $m$ is the mass of the weight, $g$ is the acceleration due to gravity and $h$ is the height or distance that the weight is lowered.

Express gain in spring potential energy

The gain in spring potential energy as the spring is elongated can be calculated using the formula $E_{sp}=\frac{1}{2}k{x}^2$, where $k$ is the spring constant, and $x$ is the distance the spring is stretched (which is the same as the distance $h$ that the weight is lowered).

Equality of energy loss and gain

Set these two energies equal to each other and solve for $h$: $mgh=\frac{1}{2}k{x}^2$. Dividing both sides by $mgh$, you will find that $1=\frac{1}{2}\frac{k{x}^2}{mgh}$. From this equation, it's clear to see that the two quantities are not equal because of the factor of $\frac{1}{2}$ in the formula for elastic potential energy, which comes from the integration when deriving the formula.

Interpretation of the result

The disparity between the loss of gravitational potential energy and the gain in spring energy is due to the way the spring stretches: it does not do so uniformly. The tension at the beginning, when the spring is not stretched, is zero and it gradually increases. The energy stored is not therefore $mgh$, but half of it.

Key concepts

Gravitational Potential Energy

Gravitational potential energy is the energy possessed by an object due to its position relative to a gravitational source, usually Earth. In simple terms, it's the energy that an object has because of its height above the ground. When we raise a weight, we do work against the force of gravity, and this work is stored as gravitational potential energy.

The formula to calculate gravitational potential energy is given by $E_{gp}=mgh$, where $m$ stands for the mass of the object, $g$ is the acceleration due to gravity (approximately 9.81 m/s^2 on Earth), and $h$ is the height the object is raised.

This concept is pivotal when discussing the exercise in question, where a weight attached to a spring loses gravitational potential energy as it is lowered. The understanding of how this energy translates into spring potential energy is crucial for students learning about energy transformation.

Hooke's Law

Hooke's Law is fundamental when studying springs and their elastic properties. It states that the force exerted by a spring is directly proportional to the displacement or stretch ($x$) from its original length, provided the limit of elasticity is not exceeded. The law is represented by the formula $F=-kx$, where $F$ is the force applied by the spring, $k$ is the spring constant, indicating the stiffness of the spring, and $x$ represents the displacement.

Understanding Hooke's Law allows us to determine the gain in spring potential energy in the provided exercise. It's the key to calculate how much energy is stored when the spring is stretched or compressed from its original position.

Energy Conservation

The principle of energy conservation is vital in physics, stating that energy cannot be created or destroyed, only transformed from one form to another or transferred from one body to another. In other words, the total energy in an isolated system remains constant over time.

This is closely related to the exercise as the gravitational potential energy lost by the weight must equal the total energy gained by the spring, discounting any work done by other forces (in an ideal scenario). The conversion factor of $\frac{1}{2}$ encountered in the solution steps is a direct result of this conservation principle, as the energy is reconfigured from gravitational potential to spring potential energy.

Spring-Mass System

A spring-mass system consists of a mass attached to a spring, which can oscillate freely. It's a classic example of a system where potential and kinetic energy are exchanged. In the static case described in the exercise, the mass attached to the spring stretches the spring until it reaches an equilibrium position, where the force of gravity pulling down on the mass is balanced by the elastic force of the spring pulling the mass upwards.

The spring's potential energy is at its maximum at this equilibrium position. As the mass is lowered and the spring stretches, the system stores energy as spring potential energy, which can be calculated using the modified Hooke’s Law in terms of energy, $E_{sp}=\frac{1}{2}k{x}^2$ where $x$ is the displacement of the mass from its natural resting position. In a dynamic situation, if the mass were to be released, it would oscillate around this equilibrium position, continually converting between its kinetic and potential energies.