Question

A variable force is given by \(F(x)=A x^{6},\) where \(A=11.45 \mathrm{~N} / \mathrm{m}^{6}\). This force acts on an object of mass 2.735 kg that moves on a frictionless surface. Starting from rest, the object moves from \(x=1.093 \mathrm{~m}\) to \(x=4.429 \mathrm{~m}\). How much does the kinetic energy of the object change?

Short answer

Answer: The change in kinetic energy of the object is approximately 13874.803 J.

Step-by-step solution

Write the given force function

The variable force acting on the object is given by: $$F(x) = A x^6$$, where A = 11.45 N/m^6.

Calculate work done by force

To calculate the work done by the variable force, we integrate the force function from the initial position to the final position: $$W = \int_{x_1}^{x_2} F(x) \, dx = \int_{1.093}^{4.429} (11.45 x^6) \, dx$$

Evaluate the integral

Let's evaluate the integral by finding the antiderivative of the force function and then using the limits of integration: $$W = \left[\frac{11.45}{7} x^7\right]_{1.093}^{4.429} = \frac{11.45}{7}(4.429^7 - 1.093^7) \approx 13874.803 \, \text{J}$$ So, the work done by the force on the object is approximately 13874.803 J.

Use the work-energy theorem

According to the work-energy theorem, the work done on the object is equal to the change in its kinetic energy: $$\Delta KE = W$$ Hence, the change in kinetic energy of the object is approximately 13874.803 J.

Key concepts

Work-Energy Theorem

Understanding the work-energy theorem is crucial when dealing with force and motion. It's a principle of physics that ties the concept of work to that of energy. In essence, the theorem states that the work done by all forces acting on an object is equal to the change in its kinetic energy. This relationship is fundamental in solving problems where forces cause objects to move.

Consider an object moving in the presence of a force; as the force does work on the object, the object’s kinetic energy changes. In the context of variable forces, like the force described by the equation \(F(x) = Ax^6\), calculating the work becomes less straightforward. This is where integral calculus comes into play. By integrating the force over the object's path, we’re effectively summing up all the tiny increments of work done along that path, resulting in the total work. When a force does positive work, the object gains kinetic energy. Conversely, if the work is negative, the object loses kinetic energy.

The work-energy theorem serves as a powerful tool for analyzing physical situations and is a foundation for understanding how forces affect motion and energy transitions in the physical world.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. Any object that's moving with speed has kinetic energy, often symbolized as \( KE \). The kinetic energy of an object depends on two factors: its mass (\( m \)) and its speed (\( v \)). The mathematical expression for kinetic energy is given by \( KE = \frac{1}{2}mv^2 \).

In our exercise, it’s important to note that the object starts from rest. This means initially, its kinetic energy is zero. As the variable force does work on the object, it begins to move and gains speed, and hence, kinetic energy. According to the problem, we need to find out how much the kinetic energy of the object changes as it moves from one point to another under the influence of the given force.

By using the work-energy theorem, we link the work done on the object to its kinetic energy change. The force does work, and as a result, the object's kinetic energy changes. This is a beautiful illustration of how forces and energy interact in a physical system, offering great insight into the dynamic nature of objects in motion.

Integral Calculus in Physics

Integral calculus is a branch of mathematics that is used extensively in physics to solve various problems, particularly those involving variable forces and moving objects. It's essential for adding up small quantities to determine a total quantity. This is exactly what we do when we calculate work done by a force over a distance when the force varies with position.

In the provided exercise, we need to calculate the work done by a force that isn't constant but changes as the object's position changes. The work done by the force is not simply the product of force and distance, because the force here is a function of position. By integrating the force function \( F(x) = Ax^6 \) with respect to position from the starting point \( x_1 \) to the endpoint \( x_2 \), we determine the total work done by the force over that interval.

Integral calculus allows us to perform this complex calculation smoothly. It's an indispensable tool in physics that enables us to analyze the full picture of dynamic processes and understand how variables related to force and motion interact over time and space. Without it, many problems in physics, including those involving variable forces, would be virtually unsolvable.