Problem 24
Question
Calculate the work required to stretch the following springs \(0.4 \mathrm{m}\) from their equilibrium positions. Assume Hooke's law is obeyed. a. A spring that requires a force of \(50 \mathrm{N}\) to be stretched \(0.1 \mathrm{m}\) from its equilibrium position b. A spring that requires 2 J of work to be stretched \(0.1 \mathrm{m}\) from its equilibrium position
Step-by-Step Solution
Verified Answer
Answer: The work required to stretch spring a is 40 J, and the work required to stretch spring b is 32 J.
1Step 1: Spring a
We will use Hooke's law for spring a:
\(F = -kx\)
\(50 \ \mathrm{N} = k \cdot 0.1 \ \mathrm{m}\)
Rearrange and solve for k:
\(k = \cfrac{50 \ \mathrm{N}}{0.1\mathrm{m}} = 500 \ \mathrm{N/m}\)
2Step 2: Spring b
(Note that we can use the formula for stored elastic potential energy in the spring, given by: W = ½kx²)
For spring b we are given W = 2 J, and x = 0.1m.
\(2\ \mathrm{J} = \cfrac{1}{2}k(0.1 \mathrm{m})^2\)
Rearrange and solve for k:
\(k = \cfrac{2 \ \mathrm{J}}{1/2 \cdot (0.1\mathrm{m})^2} = 400\ \mathrm{N/m}\)
Step 2: Calculate work (W) required to stretch springs 0.4m
We will use the formula for stored elastic potential energy in the spring again: W = ½kx²
3Step 3: Spring a
Using the values k = 500 N/m and x = 0.4 m:
\(W = \cfrac{1}{2} \cdot (500\ \mathrm{N/m} ) \cdot (0.4\ \mathrm{m})^2 = 40 \ \mathrm{J}\)
4Step 4: Spring b
Using the values k = 400 N/m and x = 0.4 m:
\(W = \cfrac{1}{2} \cdot (400\ \mathrm{N/m}) \cdot (0.4\ \mathrm{m})^2 = 32\ \mathrm{J}\)
So, the work required to stretch spring a is 40 J, and the work required to stretch spring b is 32J.
Key Concepts
Work Required to Stretch a SpringElastic Potential EnergySpring Constant CalculationSolving for Force in a Spring System
Work Required to Stretch a Spring
Stretching a spring requires doing work against the elastic force that the spring exerts. According to Hooke's Law, this force is directly proportional to the displacement from the spring's equilibrium position, and work is calculated as the integral of force over this displacement. In essence, when you stretch a spring by a displacement of 'x', the work 'W' you do is given by the equation:
\[ W = \frac{1}{2} k x^2 \]
where 'k' is the spring constant and 'x' is the displacement. This formula derives from integrating the force over the distance and represents the area under the force-displacement curve, signifying the work done. For example, stretching a spring 0.4 meters with a spring constant of 500 N/m requires 40 Joules of work. Understanding the work required is crucial for applications involving springs, such as in mechanical watches, vehicles, or even in exercise equipment.
\[ W = \frac{1}{2} k x^2 \]
where 'k' is the spring constant and 'x' is the displacement. This formula derives from integrating the force over the distance and represents the area under the force-displacement curve, signifying the work done. For example, stretching a spring 0.4 meters with a spring constant of 500 N/m requires 40 Joules of work. Understanding the work required is crucial for applications involving springs, such as in mechanical watches, vehicles, or even in exercise equipment.
Elastic Potential Energy
As a spring is stretched or compressed, it stores energy in the form of elastic potential energy. This energy is potential because it has the potential to do work when the spring is released and returns to its equilibrium position. This energy also follows the equation given by Hooke's Law:
\[ U = \frac{1}{2} k x^2 \]
The elastic potential energy 'U' is equal to the work 'W' done on the spring because energy is conserved. Hence, when you stretch a spring, not only are you doing work against the spring's force, but you are also storing energy in the spring. If the spring were released from the stretched position, it would perform an equivalent amount of work as it returns to equilibrium.
\[ U = \frac{1}{2} k x^2 \]
The elastic potential energy 'U' is equal to the work 'W' done on the spring because energy is conserved. Hence, when you stretch a spring, not only are you doing work against the spring's force, but you are also storing energy in the spring. If the spring were released from the stretched position, it would perform an equivalent amount of work as it returns to equilibrium.
Spring Constant Calculation
The spring constant 'k' quantifies the stiffness of a spring and is a fundamental parameter in Hooke's Law. To determine 'k', one must know the force applied to the spring 'F' and the displacement 'x' it produces. The formula to calculate the spring constant is:
\[ k = \frac{F}{x} \]
As shown in the exercise, if a spring requires a force of 50 N to stretch by 0.1 m, then the spring constant would be 500 N/m. Accurately identifying the spring constant is critical for designing systems where precise force responses are needed, like in engineering applications and material sciences.
\[ k = \frac{F}{x} \]
As shown in the exercise, if a spring requires a force of 50 N to stretch by 0.1 m, then the spring constant would be 500 N/m. Accurately identifying the spring constant is critical for designing systems where precise force responses are needed, like in engineering applications and material sciences.
Solving for Force in a Spring System
In a spring system, the force exerted by the spring is determined by its spring constant and the displacement from the equilibrium position. From Hooke's Law, the force 'F' is given by:
\[ F = -kx \]
The negative sign indicates that the force the spring exerts is in the opposite direction of the displacement — it's a restoring force, attempting to bring the spring back to its natural length. In applications, calculating the force is essential for understanding how the spring will react under different loads, such as in safety mechanisms, force sensors, and oscillatory systems like pendulums or vibrating strings.
\[ F = -kx \]
The negative sign indicates that the force the spring exerts is in the opposite direction of the displacement — it's a restoring force, attempting to bring the spring back to its natural length. In applications, calculating the force is essential for understanding how the spring will react under different loads, such as in safety mechanisms, force sensors, and oscillatory systems like pendulums or vibrating strings.
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