Question

How much work is required to move an object from \(x=1\) to \(x=3\) (measured in meters) in the presence of a force (in \(\mathrm{N}\) ) given by \(F(x)=2 / x^{2}\) acting along the \(x\) -axis?

Short answer

Answer: The work required is (4/3) N.m.

Step-by-step solution

Identifying given values and formula

We are given the force acting on the object \(F(x) = \frac{2}{x^2}\) and the range of movement from \(x=1\) to \(x=3\). The formula to calculate the work done by a force is given by: \(W = \int_{x_1}^{x_2} \! F(x) dx\).

Set up the integral

Now, we can set up the integral using the given force and limits of integration: \(W = \int_{1}^{3} \! \frac{2}{x^2} dx\).

Integrate the function

In order to find the value of the work done, we need to integrate the function between the given limits of integration. The integral of \(\frac{2}{x^2}\) is given by: \(\int \frac{2}{x^2} dx = -\frac{2}{x} + C\), where C is the constant of integration.

Determine the limits

Now, we need to apply the limits of integration from \(x=1\) to \(x=3\). We will use the Fundamental Theorem of Calculus to find the work done: \(W = \left[-\frac{2}{x}\right]_1^3 = -\frac{2}{3} - \left(-\frac{2}{1}\right)\)

Calculate the work done

Now, we can simply evaluate the expression to find the work done: \(W = -\frac{2}{3} + 2 = \frac{4}{3}\) So, the work required to move the object from \(x=1\) to \(x=3\) in the presence of the given force is \(\frac{4}{3}\) N.m.

Key concepts

Integration

Integration is a critical concept in calculus that allows us to find the area under curves, among other applications. In our exercise, we use integration to compute the total work done by a variable force acting on an object.
The function provided, \( F(x) = \frac{2}{x^2} \), is our force, which depends on position \(x\). To find the work done when the force varies, we integrate this force function over the desired interval (from \(x=1\) to \(x=3\)).

• Integration helps us sum infinitely small products of force and distance, which gives us the total work done.
• Without integration, calculating work for a non-constant force would be impractically complex.

Setting up the integral involves defining the limits and the function to integrate: \( W = \int_{1}^{3} \frac{2}{x^2} dx \). This process transforms our problem into solving this integral, allowing us to compute the work done effectively.

Work done

The concept of "Work done" is essential in physics and engineering. Work describes the amount of energy transferred by a force through a distance. In our case, work is determined by a force \( F(x) = \frac{2}{x^2} \) applied to move an object from \( x=1 \) to \( x=3 \).

The work done is calculated using the formula:

• \(W = \int_{x_1}^{x_2} F(x) \, dx\) - This tells us that work is the definite integral of the force function over the given limits.

By substituting \( F(x) \) and the limits into the integral, we form \( W = \int_{1}^{3} \frac{2}{x^2} \, dx \).
This process evaluates how much force is exerted on the object over the distance it moves, considering the force varies along its path.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration. It essentially tells us that to evaluate a definite integral, we need to find an antiderivative and then compute its value at the boundaries of the interval.
In our step-by-step solution, we observe this theorem in action. After finding the antiderivative of \( \frac{2}{x^2} \), given by \( -\frac{2}{x} + C \), we apply the theorem:

• The theorem states: If \( F \) is the antiderivative of \( f \) over \([a, b]\), \( \int_a^b f(x) \, dx = F(b) - F(a) \).

We compute \( \left[-\frac{2}{x} \right]_1^3 \), which involves substituting the bounds directly into this antiderivative. This calculation simplifies the process of finding the definite integral, yielding the work done as \( \frac{4}{3} \) N.m.
The Fundamental Theorem of Calculus thus provides a powerful tool to transition from indefinite integrals to actionable results.