Question

A variable force is given by \(F(x)=A x^{6},\) where \(A=16.05 \mathrm{~N} / \mathrm{m}^{6} .\) This force acts on an object of mass \(3.127 \mathrm{~kg}\) that moves on a frictionless surface. Starting from rest, the object moves from a position \(x_{0}\) to a new position, \(x=3.313 \mathrm{~m}\). The object gains \(1.00396 \cdot 10^{4} \mathrm{~J}\) of kinetic energy. What is the initial position \(x_{0}\) ?

Short answer

Based on the given information and using the work-energy theorem, the initial position of the object (𝑥₀) is approximately 2.489 meters.

Step-by-step solution

Write the integral of Force

We need to find the work done on the object by the force, which can be calculated using the integral of the force function with respect to the displacement (\(x\)). So, the work done is given by: $$ W = \int_{x_0}^{3.313} F(x) dx $$

Substitute for \(F(x)\)

We will now substitute the given force function \(F(x) = Ax^6\) into the integral: $$ W = \int_{x_0}^{3.313} Ax^6 dx $$

Integrate the function

Now we will integrate the function onto the defined interval: $$ W = \left[\frac{A}{7} x^7 \right]_{x_0}^{3.313} $$

Compute the work done

Plug in the given values for A and the final position x, and evaluate the difference to find the work done: $$ W = \frac{16.05}{7}(3.313^7 - x_0^7) $$

Use the work-energy theorem

The work-energy theorem states that the work done (W) is equal to the change in kinetic energy (∆KE). The object started from rest, so its initial kinetic energy is zero. Thus, the work done is equal to its final kinetic energy (1.00396 x \(10^4\) J). We can set up the equation as follows: $$ \frac{16.05}{7}(3.313^7 - x_0^7) = 1.00396 \cdot 10^4 $$

Solve for \(x_0\)

Now we need to solve this equation for the initial position \(x_0\). First, we will multiply both sides by \(\frac{7}{16.05}\): $$ 3.313^7 - x_0^7 = \frac{7 \cdot 1.00396 \cdot 10^4}{16.05} $$ Next, subtract \(3.313^7\) from both sides and take the seventh root to find \(x_0\): $$ x_0 = \left(\frac{7 \cdot 1.00396 \cdot 10^4}{16.05} - 3.313^7 \right)^{\frac{1}{7}} $$ By evaluating the expression, we obtain the initial position \(x_0\) (in meters): $$ x_0 \approx 2.489 \,\mathrm{m} $$

Key concepts

Work-Energy Theorem

The work-energy theorem is a fundamental concept in physics that connects the work done on an object with the change in its kinetic energy. Simply put, when work is done on an object by a force, the energy transferred by that force changes the object's kinetic energy.

Consider a scenario where an object begins at rest, meaning its initial kinetic energy is zero. When a variable force, such as the one described by the function F(x) = Ax⁶, acts on the object, it does work on the object, thus increasing its kinetic energy. In this particular problem, we know the final kinetic energy and can calculate the work done to determine the object's initial position.

The work-energy theorem can be mathematically expressed as W = ΔKE, where W represents the work done and ΔKE represents the change in kinetic energy. By calculating the integral of the force over the distance traveled, we essentially find the work done, and hence, via the work-energy theorem, can infer the changes in kinetic energy.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. It depends on both the mass of the object and the square of its velocity and is given by the formula KE = 1/2 mv², where m is the mass and v is the velocity of the object.

In the context of the given problem, the kinetic energy of the object changes because a force is acting on it, causing it to accelerate from rest. As the object moves from an initial position x₀ to a final position of x = 3.313 m, its velocity and hence kinetic energy increases.

The problem specifies that the object gains a certain amount of kinetic energy, which is directly attributable to the work done on it by the variable force. It is important to recognize that this is a clear example of the conversion of work into kinetic energy, as shown by the increase in the kinetic energy of the object.

Integral Calculus in Physics

Integral calculus is an essential tool in physics, especially when dealing with variable forces and calculating work. Integrating a force function over a path gives us the total work done by that force along that path.

In our scenario, to find the work done by the force F(x) = Ax⁶, we must integrate the function with respect to the variable x from the initial position x₀ to the final position x = 3.313 m. This integral is symbolically represented as W = ∫_x0^3.313 F(x) dx.

Once the integration is performed, we can use that to calculate the initial position of the object using the given change in kinetic energy. This application of integral calculus in physics allows us to deeply understand the relationship between the force applied, the path taken, and the resulting work done—opening the door to solving complex dynamic problems.