Problem 54 - Calculus Early Transcendentals

Question

A rigid body with a mass of \(2 \mathrm{kg}\) moves along a line due to a force that produces a position function \(x(t)=4 t^{2},\) where \(x\) is measured in meters and \(t\) is measured in seconds. Find the work done during the first 5 s in two ways. a. Note that \(x^{\prime \prime}(t)=8 ;\) then use Newton's second law \(\left(F=m a=m x^{\prime \prime}(t)\right)\) to evaluate the work integral \(W=\int_{x_{0}}^{x_{f}} F(x) d x,\) where \(x_{0}\) and \(x_{f}\) are the initial and final positions, respectively. b. Change variables in the work integral and integrate with respect to \(t .\) Be sure your answer agrees with part (a).

Step-by-Step Solution

Answer

Question: Calculate the work done by a force on a rigid body during the first 5 seconds, given that the mass of the body is 2 kg and its position function is x(t) = 4t^2. Answer: The work done during the first 5 seconds is 1600 J.

Step 1: Finding acceleration function and force function

To find the acceleration function, we'll differentiate the given position function twice with respect to time. We are given \(x(t) = 4t^2\). Then, we'll use Newton's second law to determine the force function. Newton's second law states that \(F=ma\). First derivative of position function (velocity function): \(v(t) = x'(t) = \mathrm{d}(4t^2)/\mathrm{d}t = 8t\) Second derivative of position function (acceleration function): \(a(t) = v'(t) = x''(t) = \mathrm{d}(8t)/\mathrm{d}t = 8\) Next, we'll use Newton's second law (\(F = ma\)): \(F(t) = m * a(t) = 2 * 8 = 16\) The force function is a constant, \(F(t) = 16\).

Step 2: Calculating work using the force function and integrating with respect to position

To calculate the work done, we need to evaluate the work integral \(W = \int_{x_0}^{x_f} F(x) dx\), where \(x_0\) and \(x_f\) are the initial and final positions, respectively. We need to find the initial position \(x_0 = x(0)\) and final position \(x_f = x(5)\): \(x_0 = x(0) = 4(0^2) = 0\) \(x_f = x(5) = 4(5^2) = 100\) Now, we can evaluate the work integral: \(W = \int_{x_0}^{x_f} F(x) dx = \int_{0}^{100} 16 dx = 16x\Big|_0^{100} = 16(100) - 16(0) = 1600\) The work done in the first 5 seconds using the first method is 1600 J.

Step 3: Changing variables in the work integral and integrating with respect to time

The work integral can be rewritten as \(W = \int_{t_0}^{t_f} F(t)v(t) dt\), where \(t_0\) is the initial time and \(t_f\) is the final time. In this case, \(t_0 = 0\) and \(t_f = 5\). Since we already have both \(F(t)\) and \(v(t)\), we can now compute the work integral: \(W = \int_{t_0}^{t_f} F(t)v(t) dt = \int_{0}^{5} 16(8t) dt = 128t^2\Big|_0^{5} = 128(5^2) - 128(0^2) = 1600\) The work done using the second method is also 1600 J. Therefore, the answers obtained from both methods agree, and the work done during the first 5 seconds is 1600 J.

Key Concepts

Newton's Second Law

Understanding Newton's second law is crucial to solving physics problems related to force and motion. Newton's second law states that the force acting on a body is equal to the mass of the body multiplied by its acceleration. This can be written as the equation:

• \( F = ma \)

This law explains how the motion of an object changes when it is subjected to an external force. In simple terms, it tells us that the acceleration (

• \(a\)

) of an object produced by a force (

• \(F\)

) is directly proportional to the force and inversely proportional to its mass (

• \(m\)

). This relationship allows us to calculate the force if we know the mass and the acceleration.

Position Function

A position function describes the location of a particle at various times. It is a mathematical representation that defines the path of a particle in relation to time. In the given problem, the position function is expressed as:

• \(x(t) = 4t^2\)

This quadratic equation tells us how the particle's position changes over time. The variable \(t\) represents time, and \(x(t)\) represents the position of the particle.
By examining this function, we can predict the particle's position at any given time. For example, at time \(t = 0\), the initial position \(x_0 = 0 \), and at time \(t = 5\), the final position \(x_f = 100 \). These values are essential for solving the problem of work done.

Acceleration Function

The acceleration function gives us vital information about how the velocity of an object changes over time. It is directly related to the position function through its derivatives. When you take the second derivative of the position function with respect to time, you obtain the acceleration function.
For our example, we found the acceleration function from the position function \(x(t)=4t^2\):

• First derivative (velocity function): \(v(t) = 8t\)
• Second derivative (acceleration function): \(a(t) = 8\)

This calculation shows that the acceleration is constant and equal to \(8\, \text{m/s}^2\). Knowing the acceleration is pivotal, as it connects directly to the force acting on the object via Newton's second law \(F = ma\). Understanding the acceleration function allows us to make deeper conclusions about the object's motion.

Integration in Calculus

Integration plays a crucial role in determining quantities like work in physics. When a force is applied over a certain distance, the work done is calculated by integrating the force with respect to the distance.
For the given exercise, the work done can be represented by the integral:

• \(W = \int_{x_0}^{x_f} F(x) \, dx\)

In the second approach of solving the exercise, integration is performed with respect to time, not position, using the formula:

• \(W = \int_{t_0}^{t_f} F(t) v(t) \, dt\)

This involves changing variables, transforming the integral based on the velocity function. By doing so, not only is the consistency of physical principles confirmed, but the calculation also becomes versatile.
Understanding integration helps solidify the connection between mathematical techniques and physical concepts, allowing us to compute the work done in both cases—by distance and by time—in the problem.