Question

Hooke's Law In Exercises 3-10, use Hooke's Law to determine the variable force in the spring problem. Seven and one-half foot-pounds of work is required to compress a spring 2 inches from its natural length. Find the work required to compress the spring an additional one-half inch.

Short answer

The work required to compress the spring an additional one-half inch is 5.625 inch-pounds.

Step-by-step solution

Calculate the Spring Constant \( k \)

To find the spring constant, we can rearrange Hooke's Law to \( k = \frac{F}{x} \). To find the force (\( F \)), we use the work-energy theorem which states that work done equals force times distance in the direction of the force. Hence, the force (\( F \)) in this context is the amount of work done to compress the spring 2 inches, which is 7.5 foot-pounds. Before using this value, we need to convert it to inch-pounds, since our distance is given in inches. Thus, \( F = 7.5 \times 12 = 90 \) inch-pounds. Now, \( k = \frac{F}{x} = \frac{90}{2} = 45 \) pounds per inch.

Use Hooke's Law to Calculate Additional Work

The work done to additionally compress the spring a half inch can be found using Hooke's Law, rearranged to \( W = \frac{1}{2}kx^2 \), where \( W \) is the work done, \( k \) is the spring constant, and \( x \) is the additional distance the spring is compressed. We already found \( k = 45 \) pounds per inch, and the additional compression distance (\( x \)) is a half inch, or \( x = 0.5 \) inch. Thus, the additional work done is \( W = \frac{1}{2} \times 45 \times (0.5)^2 = \frac{1}{2} \times 45 \times 0.25 = 5.625 \) inch-pounds.

Key concepts

Spring Constant Calculation

Understanding the spring constant is crucial when studying the mechanics of materials. The spring constant, denoted as 'k', represents the ratio of the force affecting the spring to the displacement caused by it. It's a measure of a spring's stiffness. The stiffer the spring, the larger the spring constant.

To calculate the spring constant using Hooke's Law, we follow the formula: \( k = \frac{F}{x} \) where 'F' is the force in pounds applied to the spring, and 'x' is the displacement in inches that this force causes. From our exercise, we know that a force of 90 inch-pounds could displace the spring by 2 inches. Hence, the spring constant is calculated to be \( 45 \) pounds per inch.

This information can be incredibly useful in understanding how much energy can be stored in a spring and how much force it can exert when released; characteristics that are crucial in a variety of engineering applications such as in automotive suspension systems and various types of machinery.

Work-Energy Theorem

The work-energy theorem is a fundamental principle in physics that states that the work done on an object is equal to the change in its kinetic energy. When applied to a spring, this theorem allows us to calculate the work needed to displace the spring from its equilibrium position.

In the context of our exercise: \( Work = Force \times Distance \) - this formula tells us that work is done when we apply a force to compress the spring over a certain distance. Our exercise had us calculate the amount of work (7.5 foot-pounds) required to compress a spring by 2 inches. The theorem helps us understand and calculate the energy transferred to the spring, thus allowing us to make predictions about the spring's potential energy and the force it can exert when it returns to its natural length.

Force and Work Relationship

Force and work are intrinsically related in the physical world, especially in the context of springs and Hooke's Law. When we speak of work in a physics sense, we're talking about the product of a force over the distance through which it acts. This relationship is essential in understanding how energy is transferred and conserved within a system.

Let's take our spring problem as an example. When we calculate the work required to compress a spring by an additional half inch, we're using the initial spring constant found earlier alongside the new displacement. Here, the work is not simply the product of force and distance but also includes the spring's resistance, which is quantified by the spring constant 'k'. So, the relationship between force and work becomes: \( W = \frac{1}{2}kx^2 \) for a spring, where 'W' is work, 'k' is the spring constant, and 'x' is the distance the spring is compressed. This formula shows us that work is also related to the properties of the spring itself (its stiffness), not just the applied force and distance.