Question

A force given by \(\vec{F}(x)=5 x^{3} \dot{x}\) (in \(\mathrm{N} / \mathrm{m}^{3}\) ) acts on a \(1.00-\mathrm{kg}\) mass moving on a frictionless surface. The mass moves from \(x=2.00 \mathrm{~m}\) to \(x=6.00 \mathrm{~m}\). a) How much work is done by the force? b) If the mass has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\) at \(x=2.00 \mathrm{~m},\) what is its speed at \(x=6.00 \mathrm{~m} ?\)

Short answer

Question: Calculate the work done by the force acting on a 1.00-kg mass and its final speed after a displacement when the force is given as a function of position, \(\vec{F}(x) = 5x^3\dot{x}\) and the mass is initially at a speed of 2.00 m/s. Answer: The work done by the force is 4550 J and the final speed of the 1.00-kg mass after displacement is 95.6 m/s.

Step-by-step solution

Calculate work done by the force

To find the work done by the force, we need to use the following formula: $$ W = \int_{x_1}^{x_2} \vec{F}(x) \cdot d\vec{x}$$ First, plug the given function for the force, \(\vec{F}(x) = 5x^3\dot{x}\), and the given limits of integration (\(x = 2.00 m\) to \(x = 6.00 m\)) into the formula. $$ W = \int_{2.00}^{6.00} (5x^3\dot{x})dx$$ Now, integrate the function with respect to the distance x: $$ W = \left[\frac{5x^4}{4}\right]_{2.00}^{6.00}$$ Substitute the values of \(x_1 = 2.00 m\) and \(x_2 = 6.00 m\) into the equation: $$ W = \frac{5(6.00)^4}{4} - \frac{5(2.00)^4}{4}$$ Calculate the work done: $$ W = 4550 J$$ The work done by the force is 4550 J.

Calculate final speed using the work-energy theorem

To find the final speed, we will use the work-energy theorem, which states: $$ W = \frac{1}{2}m(v_f^2 - v_i^2)$$ We are given the initial velocity, \(v_i = 2.00 m/s\). Now, rearrange the equation to solve for the final velocity, \(v_f\): $$ v_f^2 = \frac{2W}{m} + v_i^2$$ Plug the values for work (\(W = 4550 J\)), mass (\(m = 1.00 kg\)), and initial velocity (\(v_i = 2.00 m/s\)) into the equation: $$ v_f^2 = \frac{2(4550)}{1.00} + (2.00)^2$$ Calculate the final velocity squared: $$ v_f^2 = 9144$$ Take the square root to find the final velocity: $$ v_f = \sqrt{9144}$$ $$ v_f = 95.6 m/s$$ The final speed of the 1.00-kg mass at \(x = 6.00 m\) is \(95.6 m/s\).

Key concepts

Work Done by a Force

When studying the motion of objects, one crucial element is understanding the work done by a force. Work in physics is defined as the process of transferring energy to an object via force applied over a distance. The work done by a force is calculated by integrating the force over the distance the object travels.

In practical situations, work is done when the force has a component in the direction of the object's motion. The formula for work is given by the integral of force with respect to displacement. In our example, the force depends on the position as given by the equation \( \vec{F}(x) = 5x^3\dot{x} \). The work done can be mathematically represented as an integral, \( W = \int_{x_1}^{x_2} \vec{F}(x) \cdot d\vec{x} \), which calculates the area under the force vs. distance curve, a concept fundamental in understanding how forces impact energy transfer.

Integrating force over distance to calculate work is an application that merges concepts from physics and calculus to address practical problems.

Integration in Physics

Integration, a key operation in calculus, is used extensively in physics, especially in contexts like calculating work done by a force. Integration allows physicists to sum up infinitely many infinitesimal quantities, which is essential when dealing with continuous variables like force over a distance.

In our given problem, integration is used to find the total work done by a force that varies with its position. The integration of \( 5x^3\dot{x} \) over the distance from \( x=2.00 m \) to \( x=6.00 m \) illustrates how integration takes a variable force and calculates the cumulative effect it has along the path of an object's motion. This ability to accumulate the varying force over a continuous range is what makes integration a powerful tool in theoretical and applied physics problems alike.

Understanding how to perform integrals, such as those encountered in work calculations, is thus a foundational skill for students tackling both theoretical and experimental physics.

Kinetic Energy

Kinetic energy represents the energy an object possesses due to its motion. It can be quantified using the expression \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. When a force does work on an object, it can change the object's kinetic energy by increasing or decreasing its velocity.

In the scenario provided, we see kinetic energy at play when a 1.00-kg mass is set in motion. By calculating the work done by the force acting on this mass and utilizing the work-energy theorem, one can determine the change in the kinetic energy of the mass, which directly correlates with its change in speed. As we see from the solution, a substantial amount of work done on the mass results in a significant increase in kinetic energy, manifesting as a high final speed of the mass. Understanding kinetic energy is essential for interpreting the results of work-energy principles in physics problems.

Theoretical Physics Problems

Theoretical physics problems often incorporate complex situations that require deep understanding of physical laws and the ability to apply mathematical techniques, such as calculus, to solve these problems. In the work-energy theorem context, these problems involve the transformation of energy types and the role of forces in these transformations.

In contexts like our textbook example, where motion and energy conversion are involved, the theories and equations of physics must be manipulated mathematically to provide predictions and insights. Problems that involve calculating work done by forces or changes in kinetic energy highlight the interplay between theory and application in physics. They allow students to practice their analytical thinking and problem-solving skills in scenarios that often have real-world relevance.

Grasping theoretical physics problems requires a solid foundation in both conceptual understanding and mathematical competence—a dual mastery that is the very heart of physics education.