Problem 41 - College Physics

Question

A} 0.50-\mathrm{kg}\( mass is placed on the end of a vertical spring that has a spring constant of \)75 \mathrm{~N} / \mathrm{m}$ and eased down into its equilibrium position. (a) Determine the change in spring (elastic) potential energy of the system. (b) Determine the system's change in gravitational potential energy.

Step-by-Step Solution

Answer

(a) Change in spring potential energy: 0.16 J. (b) Change in gravitational potential energy: -0.32 J.

Step 1: Understanding the Problem

We have a vertical spring with a mass of 0.50 kg attached to it. The spring has a spring constant, \( k \), of 75 N/m. The task is to find the change in spring potential energy and the change in gravitational potential energy when the mass is eased into equilibrium.

Step 2: Apply Hooke's Law

At equilibrium, the spring will be stretched by a distance \( x \) such that the force due to gravity equals the spring force. The gravitational force \( F_g = mg \) is equal to the spring force \( F_s = kx \). Therefore, \( mg = kx \).

Step 3: Solve for Displacement (x)

Substitute the mass \( m = 0.50 \, \text{kg} \), gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \), and spring constant \( k = 75 \, \text{N/m} \) into the equation to solve for \( x \): \( 0.50 \times 9.8 = 75x \). This simplifies to \( x = \frac{0.50 \times 9.8}{75} \).

Step 4: Calculate the Displacement

Compute \( x \): \( x = \frac{4.9}{75} = 0.0653 \, \text{m} \). So, the displacement of the spring at equilibrium is approximately 0.0653 m.

Step 5: Calculate Spring Potential Energy Change

The change in spring potential energy \( \Delta U_s \) is determined using \( \Delta U_s = \frac{1}{2} k x^2 \). Substitute \( k = 75 \, \text{N/m} \) and \( x = 0.0653 \, \text{m} \): \( \Delta U_s = \frac{1}{2} \times 75 \times (0.0653)^2 \).

Step 6: Compute Spring Potential Energy Change

Calculate \( \Delta U_s \): \( \Delta U_s = \frac{1}{2} \times 75 \times 0.00426 \approx 0.16 \, \text{J} \). Therefore, the change in spring potential energy is 0.16 J.

Step 7: Calculate Gravitational Potential Energy Change

The change in gravitational potential energy \( \Delta U_g \) is given by \( \Delta U_g = -mgx \). Substitute \( m = 0.50 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( x = 0.0653 \, \text{m} \): \( \Delta U_g = -0.50 \times 9.8 \times 0.0653 \).

Step 8: Compute Gravitational Potential Energy Change

Calculate \( \Delta U_g \): \( \Delta U_g = -0.50 \times 9.8 \times 0.0653 \approx -0.32 \, \text{J} \). Therefore, the change in gravitational potential energy is approximately -0.32 J.

Key Concepts

Hooke's Law

Hooke's Law is an essential principle in understanding spring mechanics. This law tells us how the force on a spring relates to its displacement. When you apply a force to a spring, it either stretches or compresses. Hooke's Law quantifies this relationship by stating:

• Force \( F \) is proportional to the displacement \( x \).
• The formula is \( F = -kx \), where \( k \) is the spring constant.

The negative sign indicates that the force exerted by the spring opposes the direction of displacement. This is why springs tend to return to their original shape. If you hang a weight (like a 0.50 kg mass) from a spring, the force of gravity acting on the mass will equal the force exerted by the spring when it reaches equilibrium. This means \( mg = kx \), where \( mg \) (mass times gravitational acceleration) is simply the weight of the object. By rearranging the terms, we can solve for \( x \) or the displacement:\[ x = \frac{mg}{k}\]Understanding this displacement allows us to delve into potential energies, both spring and gravitational, which are crucial in this exercise.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. For objects near the Earth's surface, the formula is:

• \( U_g = mgh \)
• Where \( m \) is mass, \( g \) is the acceleration due to gravity (9.8 \( m/s^2 \)), and \( h \) is the height above a reference point.

When a 0.50 kg mass is placed on a spring, it moves downward and loses height. As it descends, it loses gravitational potential energy. In this context, we calculate the change in GPE as:\[ \Delta U_g = -mgx\]The negative sign indicates a decrease in gravitational potential energy as the mass moves downward. For our example, based on the displacement \( x = 0.0653 \) m, the change in GPE is approximately -0.32 J. This energy change accompanies the transformation of energy from the gravitational field into spring potential energy.

Spring Potential Energy

Spring potential energy, also known as elastic potential energy, is the stored energy in a spring when it is either compressed or stretched. The formula for calculating the spring's potential energy is given by:

• \( U_s = \frac{1}{2}kx^2 \)
• Where \( k \) is the spring constant, and \( x \) is the displacement from equilibrium.

Consider the vertical spring example. When the 0.50 kg mass stretches the spring to its new equilibrium position after being eased down, energy is stored in the spring.The spring potential energy change can be found using:\[ \Delta U_s = \frac{1}{2}k x^2\]For a spring constant \( k \) of 75 N/m and a displacement \( x \) of 0.0653 m, the calculation yields a change of approximately 0.16 J in spring potential energy. This increase is a result of the energy transferred from the gravitational potential energy as the spring stretches.