Question

Multiple Choice A force of \(F(x)=350 x\) newtons moves a particle along a line from \(x=0 \mathrm{m}\) to \(x=5 \mathrm{m}\) . Which of the following gives the best approximation of the work done by the force? (A) 1750 \(\mathrm{J} \quad\) (B) 2187.5 \(\mathrm{J}\) (C) 2916.67 \(\mathrm{J}\) (D) 3281.25 \(\mathrm{J} \quad\) (E) 4375 \(\mathrm{J}\)

Short answer

The correct answer is (E) 4375 J

Step-by-step solution

Define the integral

In order to calculate the total work done by the force, calculate the definite integral of the force \(F(x) = 350x\) from 0 to 5. Thus, the integral to be solved is \(\int_{0}^{5} 350x dx\).

Compute the integral

Now compute the integral. The antiderivative of \(350x\) is \(350 \frac{x^2}{2}\). Plug the limits 0 and 5 into the antiderivative: \(350 \frac{5^2}{2} - 350 \frac{0^2}{2}\).

Simplify expression

Simplify the expression to get the final answer. The right part of the subtraction is zero, so it becomes, \(350 \frac{25}{2}\), which is equal to 4375.

Key concepts

Definite Integral

The concept of a definite integral plays a pivotal role in calculus, and it is essential in understanding work done by force. In a practical sense, when you calculate the work done by a force along a displacement, you are essentially finding the area under the curve of the force function over a certain interval.

This is achieved through the process of integration. If the force function is denoted by \( F(x) \), and the object moves from point \( a \) to point \( b \), the work done is represented by the definite integral \( \int_{a}^{b} F(x) \, dx \). The 'dx' shows that integration occurs with respect to the variable x, which typically represents the distance or displacement.

Using the definite integral to calculate work allows the consideration of variable forces, where the force might change as the object moves. For instance, in the textbook example, \( F(x) = 350x \) Newtons changes as \( x \) changes, and the integration from \( x=0 \) meters to \( x=5 \) meters accurately accounts for this variation to calculate the total work done.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the original function that we started with. In our example, if \( F(x) = 350x \), then an antiderivative of \( F \) might take the form \( G(x) = 175x^2 + C \), where \( C \) represents a constant.

Why is this concept important for work calculations? To find the work done by the force over a displacement, we must evaluate the definite integral of the force function, which involves finding the antiderivative first. The process involves taking the original function \( F(x) \) and determining another function \( G(x) \) such that \( G'(x) = F(x) \).

After finding the antiderivative, the second step is to apply the Fundamental Theorem of Calculus, which links the definite integral of a function to its antiderivative. For our work calculation, we use the antiderivative to evaluate the definite integral at the bounds \( a \) and \( b \), subtracting one from the other to find the total work done.

Force and Motion in Calculus

Force and motion are foundational concepts in mechanics that are elegantly described by calculus. Newton's second law of motion, \( F = ma \), states that force is the product of mass and acceleration. In terms of calculus and particularly in the context of variable forces, mass \( m \) is generally a constant, and acceleration \( a \) is the second derivative of the position function \( s(t) \), with respect to time \( t \), or \( a = s''(t) \).

To connect force with the change of position, we often integrate the acceleration twice to find the position function. Similarly, working backward, if we're given a force function dependent on position, \( F(x) \), we can find the work done by integrating this force function over the path of the object.

Understanding the relationship between force and motion through integration allows us to estimate the work done over a path even when the force varies with position, contributing a much more precise and dynamic understanding of mechanics than would otherwise be possible with constant forces.

Newton's Laws in Calculus

Applying Newton's Laws within the framework of calculus provides a powerful tool to solve problems involving force and motion. Newton's laws describe the relationship between a body and the forces acting upon it, and the body's motion in response to those forces.

Let's break it down with an example relevant to our textbook problem. According to Newton's first law (the law of inertia), a particle remains at rest or in uniform motion unless acted upon by an external force. This law implies that the net work done when a particle remains steady or moves with constant velocity is zero, as there is no change in kinetic energy.

With Newton's second law, which at its core is a differential equation, we can describe the changes in motion caused by varying forces. Finally, Newton's third law (action-reaction) helps us understand that forces always occur in pairs, which can be useful when considering conservation of energy or momentum in a system.

Calculus enables us to apply these laws over real-world variables that are often in flux. For instance, when calculating the work done by a varying force as shown in our problem, we utilize these laws as interpreted through integrals and derivatives to come to a solution that encompasses the full scope of the motion and force interaction.