Question

An \(x\) -component of a force has the dependence \(F_{x}(x)=-c x^{3}\) on the displacement \(x\), where the constant \(c=19.1 \mathrm{~N} / \mathrm{m}^{3} .\) How much work does it take to oppose this force and change the displacement from \(0.810 \mathrm{~m}\) to \(1.39 \mathrm{~m} ?\)

Short answer

Question: Calculate the work done to move against a force with an x-component given by \(F_x(x) = -19.1 x^3\) from an initial displacement of \(0.810 \ \text{m}\) to a final displacement of \(1.39 \ \text{m}\). Answer: The work done to move against the force is approximately 23.312 J.

Step-by-step solution

Identify the given information.

We are given: - The force's x-component dependence on displacement: \(F_x(x) = -c x^3\) - The constant \(c = 19.1 \ \text{N/m}^3\) - Initial displacement \(x_1 = 0.810 \ \text{m}\) - Final displacement \(x_2 = 1.39 \ \text{m}\)

Write the equation for work done.

The work done by the force when moving against the force from \(x_1\) to \(x_2\) is given by the integral of the force with respect to the displacement: \(W = \int_{x_1}^{x_2} F_x(x) \ \text{d}x\)

Substitute the force's x-component.

Replace \(F_x(x)\) with the given equation \(-c x^3\): \(W = \int_{x_1}^{x_2} (-c x^3) \ \text{d}x\)

Substitute the given values.

Replace \(c\) with its given value \(19.1 \ \text{N/m}^3\), \(x_1\) with \(0.810 \ \text{m}\), and \(x_2\) with \(1.39 \ \text{m}\): \(W = \int_{0.810}^{1.39} (-19.1 x^3) \ \text{d}x\)

Integrate the expression.

Integrating the expression gives: \(W = -19.1 \int_{0.810}^{1.39} x^3 \ \text{d}x = -19.1 \left[\frac{1}{4} x^4 \right]_{0.810}^{1.39}\)

Evaluate the integral.

Evaluate the integral at the limits, and then calculate the final value: \(W = -19.1 \left[\frac{1}{4} (1.39^4 - 0.810^4)\right]\) \(W ≈ -19.1 \times 1.2199\) \(W ≈ -23.312 \ \text{J}\) Since the work done should oppose the force, the magnitude of the work done is \(23.312 \ \text{J}\).

Key concepts

Force-Displacement Relation

Understanding the force-displacement relation is essential for comprehending how forces do work in physics. Work is done when a force causes a displacement of an object. The general formula for work (\( W \)) is derived from the force applied (\( F \times \text{displacement} \times \text{cos}(\theta)\times \text{displacement} \(\text{cos}(\theta)\text{displacement} \times \text{displacement} \text{displacement} \times \text{displacement} \(\text{displacement} \times \text{displacement} \(\text{displacement} \times \text{displacement} \(\text{displacement} \text{displacement} \(\text{displacement} \bra{displacement} \text{displacement} \bra{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text{displacement} \text:

Integration in Physics

Integration is a powerful mathematical tool used in physics to combine small pieces of information to calculate quantities like work, area, or volume. The case where force varies with displacement, which often happens in real-world physics problems, requires integrating the force over the path of movement to find the total work done.

In the given problem, the integration process is applied to the function \force with respect to the displacement. Integration finds the accumulated effect of force over the interval from initial displacement \(x_1\force acts. This variable force poses a unique challenge as it cannot be calculated using simple multiplication.

The integral symbol, \(\) or \(\) represents this summation of forces at infinitely many points between \(x_1\br>The definite integral, which is denoted by upper and lower limits on the integral sign, is akin to summing the work done over tiny increments of displacement, eventually providing the total work done over the specified range. This is why Step 5 in the solution involves integrating the force function from the initial to final positions.

Work-Energy Principle

The work-energy principle is a cornerstone concept in physics that connects the work done by forces to changes in energy. According to this principle, work done on an object is equal to the change in its kinetic energy. This is particularly useful when forces are conservative, meaning they have a potential energy associated with them.

In simple terms, if you use force to push an object, thereby making it move, you're transferring energy to that object; the object’s kinetic energy increases as a result. Meanwhile, if a force opposes motion, it can be seen as removing energy from the object, resulting in a decrease in kinetic energy.

In the context of the exercise, work is done against the force, which means we're opposing the force and decreasing the kinetic energy of the system. Therefore, the work done has a negative value because the force is in the opposite direction of the displacement. However, the actual magnitude of work done remains positive, as we're interested in the amount of energy transferred, regardless of direction. The final result reflects the energy required to move the object from one position to another against the varying force.