Problem 223 - Calculus Volume 1

Question

For the following exercises, find the work done. What is the work done moving a particle from \(x=0\) to \(x=1 \mathrm{m}\) if the force acting on it is \(F=3 x^{2} \mathrm{N} ?\)

Step-by-Step Solution

Answer

The work done is 1 Joule.

Step 1: Understand the Formula for Work

The work done by a force over a displacement is given by the integral of the force over the path of the displacement. For a variable force, the work done is expressed as \( W = \int F(x) \, dx \). Our given force is \( F(x) = 3x^2 \) Newtons, and we're moving from \( x=0 \) to \( x=1 \) meter.

Step 2: Set Up the Integral

To find the work done, we'll set up the integral of the force function. This means we need to evaluate \( \int_{0}^{1} 3x^2 \, dx \).

Step 3: Integrate the Function

Calculate the integral of the function. The integral of \( 3x^2 \) with respect to \( x \) is \( x^3 \). Thus, \[ \int 3x^2 \, dx = [x^3] \].

Step 4: Evaluate the Integral with Limits

Plug in the limits of integration into the antiderivative found. This yields \[ W = [1^3] - [0^3] = 1 - 0 = 1 \text{ J} \].

Step 5: Conclusion

The calculated work done when moving the particle from \( x=0 \) to \( x=1 \) meter is \( 1 \text{ Joule} \). This comes from applying the limits to the integrated force function.

Key Concepts

Variable Force Integration

When you're dealing with forces that change with position, such as the force in this exercise, it's crucial to understand variable force integration. Variable force means that the force applied to an object isn't constant and varies based on position or time. For forces like these, calculating the amount of work requires breaking down the problem using calculus.
The work done (W) by a force over a distance is found through integration. Essentially, you integrate the force function with respect to the distance over which the force is applied:\[ W = \int F(x) \, dx \]

• This integral encompasses the total area under the force curve as the object moves along a specified path.
• During integration, the variable force function is mathematically "accumulated" over a specific displacement or interval, providing the total work done.

Variable force integration transforms the complex problem of calculating work with a non-constant force into a tangible solution. Understanding how to set up and solve these integrals is key to mastering work calculations involving variable forces.

Integral Calculus

Integral calculus is the mathematical approach we use to determine the total value accumulated over a continuous change. It's fundamental in physics, especially when dealing with variable forces. With integral calculus, you can precisely calculate the work done on an object when the force applied varies with distance.
In this exercise, we see the integral calculus equation in action:

• The force function, given as \( F(x) = 3x^{2} \), is integrated over the range from \( x = 0 \) to \( x = 1 \) meter.
• The antiderivative of \( 3x^{2} \) is found to be \( x^{3} \).

After finding the antiderivative, you evaluate it over the specified limits to determine the work done:\[ W = [x^{3}]_{0}^{1} = 1^{3} - 0^{3} = 1 \, \text{J} \]This process clearly shows how integral calculus allows us to sum up the effects of variable forces, yielding the total work done as the particle moves along the path. It makes handling complex problems manageable and is an essential tool in physics.

Displacement in Physics

In physics, displacement refers to the change in an object's position from its initial point to its final point. It's a vector quantity, meaning it has both magnitude and direction, and it is crucial when computing work, as work is performed over a displacement.
For this particular exercise:

• The displacement is from \( x = 0 \) to \( x = 1 \) meter, indicating a one-directional movement along the x-axis.
• The work calculation fundamentally depends on this displacement, as it determines the limits for the integral of the variable force.

The concept of displacement is not just the distance traveled but the specific position change along the path. That’s why the limits in the integral are critically important. They define the actual physical movement used in calculating the work done by a force applied over a distance. Recognizing and correctly applying the concept of displacement in problems ensures accurate and meaningful solutions in physics calculations.