Problem 50 - Calculus Early Transcendentals

Question

A 10-kg mass is attached to a spring that hangs vertically and is stretched 2 m from the equilibrium position of the spring. Assume a linear spring with \(F(x)=k x\) a. How much work is required to compress the spring and lift the mass 0.5 m? b. How much work is required to stretch the spring and lower the mass 0.5 m?

Step-by-Step Solution

Answer

Answer: a. The work required to compress the spring and lift the mass 0.5 m is -42.79 J. b. The work required to stretch the spring and lower the mass 0.5 m is 55.18 J.

Step 1: Identify the variables

Given: Mass of the object (m) = 10 kg Initial stretch of the spring (x1) = 2 m Force equation for a linear spring: F(x) = kx We need to find the work required to: a. Compress the spring and lift the mass 0.5 m (x2 = 1.5 m) b. Stretch the spring and lower the mass 0.5 m (x2 = 2.5 m)

Step 2: Calculate the spring constant (k)

In order to determine the spring constant, we can use Hooke's law and the equation for the gravitational force acting on the mass: \(F_{spring} = kx_1 = mg\) Here, g is the gravitational acceleration (9.81 m/s²). Solving for k, we get: \(k = \frac{mg}{x_1} = \frac{10\ kg \times 9.81\ m/s^2}{2\ m} = 49.05\ N/m\)

Step 3: Determine the initial potential energy (PE1)

At the initial position (x1), the potential energy stored in the spring is: \(PE_1 = \frac{1}{2}kx_1^2 = \frac{1}{2} \times 49.05\ N/m \times (2\ m)^2 = 98.1\ J\)

Step 4: Calculate the work done for both cases a and b

a. Case when the spring is compressed and the mass is lifted (x2 = 1.5 m) The potential energy at position x2 is: \(PE_2 = \frac{1}{2}kx_2^2 = \frac{1}{2} \times 49.05\ N/m \times (1.5\ m)^2 = 55.31\ J\) The work done (W1) to compress the spring and lift the mass from x1 to x2 is: \(W_1 = PE_2 - PE_1 = 55.31\ J - 98.1\ J = -42.79\ J\) b. Case when the spring is stretched and the mass is lowered (x2 = 2.5 m) The potential energy at position x2 is: \(PE_2 = \frac{1}{2}kx_2^2 = \frac{1}{2} \times 49.05\ N/m \times (2.5\ m)^2 = 153.28\ J\) The work done (W2) to stretch the spring and lower the mass from x1 to x2 is: \(W_2 = PE_2 - PE_1 = 153.28\ J - 98.1\ J = 55.18\ J\)

Step 5: Present the results

a. The work required to compress the spring and lift the mass 0.5 m is -42.79 J. b. The work required to stretch the spring and lower the mass 0.5 m is 55.18 J.

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle in mechanics, describing how springs behave when forces are applied. It states that the force needed to extend or compress a spring by some distance is proportional to that distance. The formula for Hooke's Law is expressed as:\[ F(x) = kx \]where:

• \( F(x) \) is the force exerted by the spring in newtons (N),
• \( k \) is the spring constant in newtons per meter (N/m),
• \( x \) is the displacement from the spring's equilibrium position in meters (m).

This law only holds within the elastic limit of the spring, meaning that the force applied should not cause permanent deformation. Understanding this law helps us predict how much potential energy a spring can store at a given displacement, as well as the work required to stretch or compress it by a certain amount.

Spring Constant

The spring constant, represented by \( k \), is a measure of a spring’s stiffness. It quantifies the amount of force necessary to modify the spring's length by one meter. The equation for finding \( k \) is derived from Hooke's Law:\[ k = \frac{F}{x} \]where:

• \( F \) is the force applied,
• \( x \) is the displacement caused by the force.

In the exercise, the spring constant was calculated using the gravitational force acting on a mass (10 kg) and its displacement (2 m). By rearranging the relationship \( F_{spring} = mg \equiv kx_1 \), we found that:\[ k = \frac{mg}{x_1} \approx 49.05 \text{ N/m} \]This spring constant indicates the energy stored in the spring per unit of length displaced from its equilibrium, reflecting the spring's resistance to being compressed or stretched.

Potential Energy

Potential energy in the context of springs is the energy stored within the spring when it is either compressed or stretched. This can be calculated using the formula:\[ PE = \frac{1}{2}kx^2 \]where:

• \( PE \) denotes potential energy in joules (J),
• \( k \) is the spring constant,
• \( x \) is the displacement from its equilibrium position.

In the given exercise, the potential energy was calculated for different positions of the spring:

• Initially at 2 m, \( PE_1 = 98.1 \text{ J} \),
• Compressed to 1.5 m, \( PE_2 = 55.31 \text{ J} \),
• Stretched to 2.5 m, \( PE_2 = 153.28 \text{ J} \).

Potential energy decreases when the spring is compressed and increases when it is stretched. This concept is crucial for understanding the work needed to change a spring's length, and how energy is conserved within mechanical systems.

Gravitational Force

Gravitational force is the force of attraction acting between any two masses. On Earth's surface, it provides a weight force that pulls objects towards the center of the planet. The gravitational force can be calculated using the equation:\[ F_g = mg \]where:

• \( F_g \) is the gravitational force in newtons (N),
• \( m \) is the mass in kilograms (kg),
• \( g \) is the acceleration due to gravity, approximately 9.81 m/s².

In this exercise, the gravitational force acting on a 10 kg mass is calculated as:\[ F_g = 10 \text{ kg} \times 9.81 \text{ m/s}^2 = 98.1 \text{ N} \]This force is key in determining the spring constant when balancing against spring force. Additionally, it influences the work required to lift or lower the mass as the gravitational force acts against movements. Understanding gravitational force helps explain how energy is transferred and how to manage these forces in practical scenarios.