Problem 56 - Calculus Early Transcendentals

Question

A rigid body with a mass of 2 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 \(5 \mathrm{s}\) 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 \(W=\int_{x_{0}}^{x_{1}} F(x) d x,\) 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 on a 2 kg particle in the first 5 seconds after it starts moving along a straight line according to the position function x(t) = 4t^2. Answer: The work done on the particle in the first 5 seconds is 1600 J.

Step 1: Calculate the acceleration

To find the acceleration function, we have to find the second derivative of the position function with respect to time i.e., \(a(t) = x^{\prime\prime}(t)\). The position function is given as \(x(t) = 4t^2\). #a(t) = x^{\prime\prime}(t) = \frac{d^2}{dt^2}(4t^2) = 8#

Step 2: Calculate the force function

Now we will use Newton's second law: \(F = ma = m\times a(t)\) \(F(t) = m \cdot a(t) = 2 \cdot 8 = 16\)

Step 3: Calculate the initial and final positions

The position function is given as \(x(t) = 4t^2\). The initial position will be calculated by substituting t=0, and the final position will be calculated by substituting t=5. Initial Position: \(x_0 = x(0)= 4 \cdot 0^2 = 0\) Final Position: \(x_1 = x(5) = 4 \cdot 5^2 = 100\)

Step 4: Integrate the force function and find the work done

To find the work done, we will integrate the force function with respect to the position: \(W = \int_{x_0}^{x_1} F(x) dx\) Since F(t) is constant (16), and x(t) = 4t^2, \(W = \int_{0}^{100} 16 dx = 16x \Big|_0^{100} = 16(100) - 16(0) = 1600 J\) #b: Changing variables and integrating with respect to time#

Step 1: Change variables in the work integral

We know \(W=\int_{x_{0}}^{x_{1}} F(x) d x\). Now, let's change variables by multiplying and dividing by \(dt\). Thus: \(W=\int_{0}^{5} F(x) ~\frac{dx}{dt}~ dt\) We have \(F(t) = 16\) and \(x'(t) = \frac{dx}{dt} = 8t\)

Step 2: Integrate with respect to time

Substitute the values of F(t) and x'(t) in the above equation and integrate: \(W = \int_{0}^{5} 16 \cdot 8t dt = 128 \int_{0}^{5} t dt = 128 \cdot \frac{1}{2}t^{2}\Big|_0^{5} = 64(5^2) - 64(0^2) = 1600 ~J\) In both parts (a) and (b), we obtain the same answer for the work done, which is 1600 J, thus our calculations are consistent.

Key Concepts

Newton's Second Law

Understanding Newton's Second Law is crucial when calculating work done in physics. It states that the force applied on an object is equal to the rate of change of its momentum, which, for a constant mass, simplifies to the product of the mass and the acceleration of the object (\( F = ma \)). In the case of the rigid body moving along a line, we used this law to relate the force exerted on the body to its mass and acceleration.

Since the body has a constant mass (\( 2 kg \)), and acceleration (\( a = 8 m/s^2 \)), applying Newton's second law gives us a constant force (\( F = ma = 2 \times 8 = 16 N \)). This constant force simplifies our work calculation because the force doesn't change with position or time, making the integration process straightforward. Without understanding Newton's Second Law, the connection between the body's motion and the force applied would remain unclear.

Calculus Integration

In physics, calculus integration is often required to solve problems related to motion, such as calculating work done. Work is defined as the integral of force over displacement. This is where integration becomes essential. You can think of integration as a way to add up tiny pieces of work done over every infinitesimal part of the path the object travels.

For the given exercise, we performed an integration to find the work done by a constant force as the body moves from its initial position to its final position. In mathematical terms, we computed the definite integral of a constant function, which is a fundamental application of integration in calculus. Had the force varied with position, we would have needed to perform a more complex integration, highlighting the power and necessity of calculus in solving real-world physics problems.

Force and Acceleration

The concept of force and acceleration is intertwined in mechanics. Acceleration is how fast an object's velocity changes with time, and force is any interaction that, when unopposed, changes the motion of an object. Newton's Second Law bridges the gap between these two concepts by quantifying the relationship.

For the given textbook problem, we examined the simple scenario of constant force and constant acceleration. The body's acceleration is determined by taking the second derivative of the position function with respect to time, which reveals that the acceleration is not dependent on time. With acceleration known, the constant force exerted on the object is straightforward to calculate. This directly affects the work done on the object, as work is the result of a force causing displacement. Understanding the dynamics of force and acceleration is critical for predicting the work required to move objects and for analyzing various mechanical systems.