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, working against the force, from \(0.730 \mathrm{~m}\) to \(1.35 \mathrm{~m} ?\)

Short answer

Answer: Approximately 57.92 Joules of work is needed to change the displacement from 0.730 m to 1.35 m against the given force function.

Step-by-step solution

Set up the integral

To find the work done against the force, we need to integrate the force function over the interval from the initial displacement \(x_i = 0.730\,\text{m}\) to the final displacement \(x_f = 1.35\,\text{m}\): $$ W = -\int_{x_i}^{x_f} F_x(x) \,\mathrm{d}x $$ Here, we're using a negative sign because we want to find the work done against the force (opposite to the force's direction).

Substitute the given force function

Now, substitute \(F_x(x)=-k x^4\) into the integral and plug in the constant \(k=20.3\,\text{N/m}^4\): $$ W = -\int_{0.730}^{1.35} (- 20.3 x^4) \,\mathrm{d}x $$

Integrate the function

Next, we integrate the function with respect to x: $$ W = -\left[-\frac{20.3 x^5}{5} \Bigg|_{0.730}^{1.35}\right] $$

Evaluate the definite integral

Now, we evaluate the integral at both the upper and lower limits: $$ W = -\left[-\frac{20.3 \cdot (1.35)^5}{5} + \frac{20.3 \cdot (0.730)^5}{5}\right] $$

Calculate the work

Finally, we calculate the work by solving the expression: $$ W \approx 57.92\,\text{J} $$ So, it takes around 57.92 Joules of work to change the displacement from \(0.730\,\text{m}\) to \(1.35\,\text{m}\) against the force.

Key concepts

Force-Displacement Relation

Understanding the force-displacement relation is fundamental in physics, particularly when dealing with the concept of work. Work relates to how a force applied over a certain distance can move an object or oppose such movement.

In the provided exercise, we're dealing with a force that varies with displacement according to the equation given as: $$F_{x}(x)=-k x^{4}$$
This means that the force is not constant but instead changes as the object moves, which is a common scenario in real-world problems. This is why it’s crucial to know what happens at every point along the path of the object, hence the necessity to integrate over a range of displacements - the distance the object is moved.

The power of the force-displacement relation lies in its ability to provide a complete description of how forces affect movement along a path, enabling us to calculate physical quantities like work done, which is essentially the area under the force-displacement curve.

Definite Integral in Physics

The definite integral is a key mathematical tool in physics, used to determine the total accumulation of a quantity, such as work or charge, over a specific interval. When you see an integral sign, think of it as adding up tiny bits of something over a certain range.

In our scenario, we’re interested in how much work is done against a variable force over a particular displacement range. This is a classic case where a definite integral becomes indispensable. The calculation:$$W = -\frac{20.3}{5} \bigg[ (1.35)^5 - (0.730)^5 \bigg]$$converts the continuous and varying force over a path into a single value of work. It’s crucial to have an accurate integral over the interval from the initial to the final position, as the sum of the work done at each infinitesimal displacement gives us the total work required against the force.

Work-Energy Principle

The work-energy principle is a cornerstone of classical mechanics, linking the work done by or against forces to changes in energy. It tells us that the total work done on an object is equal to the change in its kinetic energy. This principle can be particularly useful when forces are not constant, as the integration of force over the displacement gives us the work done.

In the given exercise, we use this principle to calculate the work done against the force. Since the force is expressed as a function of its displacement, we integrate the force over a displacement interval to find the total work. The result, which turned out to be approximately 57.92 Joules, represents the energy that must be provided to overcome the force and move the object from one position to another.

It's important to highlight that work and energy are scalar quantities, and they are traditionally measured in Joules in the SI unit system. The work-energy principle provides us with a powerful way to analyze many physical situations without dealing directly with the complex forces in action.