Question

A force has the dependence \(F_{x}(x)=-k x^{4}\) on the displacement \(x\), where the constant \(k=20.3 \mathrm{~N} / \mathrm{m}^{4}\). How much work does it take to change the displacement from \(0.73 \mathrm{~m}\) to \(1.35 \mathrm{~m} ?\)

Short answer

Answer: The work done is approximately \(-1.235 J\).

Step-by-step solution

Write down the given information and the formula for work

We have the force function \(F_x(x) = -k x^4\), with the constant \(k = 20.3 N/m^4\). We need to find the work done when the displacement changes from \(0.73m\) to \(1.35m\). The formula for calculating work is given by the integral of the force function with respect to displacement: $$ W=\int_{x_1}^{x_2} F_x(x) dx $$

Substitute the given values into the formula

Substitute the given values of force function, initial displacement (\(x_1 = 0.73m\)), and final displacement (\(x_2 = 1.35m\)) into the formula for work: $$ W = \int_{0.73}^{1.35} -20.3 x^4 dx $$

Integrate the force function with respect to displacement

Now, integrate the force function: $$ W = -20.3 \int_{0.73}^{1.35} x^4 dx $$ $$ W = -20.3[\frac{1}{5}x^5]_{0.73}^{1.35} $$

Evaluate the integral at the limits

Substitute the limits of integration and find the difference: $$ W = -20.3[\frac{1}{5}(1.35^5) - \frac{1}{5}(0.73^5)] $$

Calculate the work done

Perform the calculations to find the value for the work done: $$ W \approx -20.3(0.0646 - 0.00375) = -20.3(0.06085) \approx -1.235 \mathrm{J} $$ The work done to change the displacement from \(0.73m\) to \(1.35m\) is approximately \(-1.235 J\). Since the work done is negative, it indicates that the force acts against the direction of the displacement.

Key concepts

Work-Energy Principle

At the heart of classical physics lies the work-energy principle, which relates the work done on an object to the change in its kinetic energy. In the simplest form, the work done by a constant force acting on an object is calculated as the product of the force and the displacement in the direction of the force.

However, when the force is variable as in our exercise's problem statement, the principle gets more nuanced. The work-energy theorem states that the work done on a system by all external forces equals its change in kinetic energy. Mathematically, this is represented as \( W = \Delta KE \) or \( W = KE_{final} - KE_{initial} \).

In the provided exercise, we see a specific example where a force is dependant on the displacement in a non-linear way, thus requiring the use of calculus to compute the work done across a displacement. Recognizing that the work done over a variable force can manifest as changes in different forms of energy, not just kinetic energy, is crucial for understanding the full extent of this principle. It's an essential concept that bridges the gap between forces and energy, two of the foundational pillars in physics.

Integration in Physics

Integration is a mathematical tool that plays a pivotal role in physics, particularly when dealing with variables that change continuously. Integration can be thought of as the process of adding up infinitesimally small quantities to find a total amount. This concept is crucial in calculating work done by a variable force over a displacement.

In our example with the force function \( F_x(x) = -k x^4 \), the use of integration is necessary because the force varies depending on the displacement. Unlike simple problems where force is constant, the work done cannot be figured out by just multiplying force by displacement. Instead, we compute the integral of the force over the range of displacement, which essentially sums up the tiny bits of work done over each infinitesimally small piece of the path.

Mathematical Interpretation of Work

Specifically, the work done is the integral of force with respect to displacement, given by the equation \( W=\int_{x_1}^{x_2} F_x(x) dx \) where \( x_1 \) and \( x_2 \) represent the initial and final positions. The solution steps show how integration is executed, demonstrating that it is the area under the force-displacement graph, which represents the total work done on the object.

Force-Displacement Relationship

The nature of the relationship between force and displacement is central to predicting the work done in systems where the force does not remain constant. When a force varies with displacement, it is described by a force-displacement function, such as \( F_x(x) = -k x^4 \) in the text's exercise.

In such cases, the work cannot be simply calculated as the product of force and displacement. Here, the force changes as the object moves, and so the tiny amounts of work done at each small increment of displacement must be considered. This is the essence of the integral calculus application in physics.

• The variable \(x\) represents the displacement and acts as the independent variable in the force function.
• The coefficient \(k\) reflects the constant of proportionality that characterizes the relationship between the force and the displacement in the given scenario.

Understanding this relationship is pivotal for correctly applying the concept of work. As seen in the solved exercise, by integrating the force function with respect to displacement, you calculate the work done on the object over the specified interval. The sign of the work (positive or negative) also provides insight into the direction of the force in relation to the direction of displacement - in our case, a negative work signifies that the force acts against the direction of displacement.