Question

A particle of mass \(m\) is subjected to a force acting in the \(x\) -direction. \(F_{x}=(3.0+0.50 x) \mathrm{N}\). Find the work done by the force as the particle moves from \(x=0\) to \(x=4.0 \mathrm{~m}\)

Short answer

Answer: The work done by the force is \(16 \, Joules\).

Step-by-step solution

1. Set up the integral expression for the work done.

To find the work done by the force, write down the integral expression for the work done over the interval \([0, 4]\): \(W = \int_{0}^{4} F(x) dx = \int_{0}^{4} (3.0 + 0.50x) dx\).

2. Evaluate the integral.

Now, evaluate the integral by finding the antiderivative of the force function and evaluating at the bounds. To find the antiderivative of \(3.0 + 0.50x\), we have: \(\int (3.0 + 0.50x) dx = 3.0x + 0.25x^2 + C\), where \(C\) is the constant of integration. As we are evaluating this between definite limits, the constant \(C\) has no effect on our calculation. Now, evaluate at the bounds: \(W = [3.0x + 0.25x^2]_0^4 = (3.0(4) + 0.25(4^2)) - (3.0(0) + 0.25(0^2))\).

3. Calculate the work done.

Calculate the work done by simplifying the expression: \(W = (12 + 0.25(16)) - (0 + 0) = 12 + 4 = 16 J\). So the work done by the force as the particle moves from \(x = 0\) to \(x = 4.0 m\) is \(16 \, Joules\).

Key concepts

Force in Physics

In the realm of physics, force is a fundamental concept that describes an interaction that causes an object to change its state of motion. Force can cause a stationary object to move or change the velocity of a moving object. In our exercise, the force assigned to the particle is a variable force; it depends on the position of the particle along the x-axis.

Force can be expressed mathematically. In this case, it is given by the equation:

• \( F(x) = 3.0 + 0.50x \)

This equation means the force has two components:

• A constant force of \(3.0 \, N\)
• A variable component that increases with x as \(0.50x \, N\)

The particle moves due to this force, highlighting how forces in physics can be position-dependent, not just constant.

Understanding force is crucial because it is the foundation for concepts like work and energy. Force is a vector, which means it has both magnitude and direction. In the exercise, the force is directed along the x-axis, simplifying our calculations. Recognizing these nuances makes solving problems in physics more intuitive.

Integration in Calculus

Integration is a powerful mathematical tool used extensively in physics to calculate quantities like work when a force varies with position. The work done by a force can be calculated using the integral of the force over the displacement in question. In simpler terms, integration allows us to add up small contributions of force over a path to find total work.

In the given textbook problem, the work (\( W \)) done by a variable force is found through integrating the force function:\[W = \int_{0}^{4} (3.0 + 0.50x) \, dx\]This integration assesses the total effect of the variable force as the particle moves from 0 to 4 meters.

• Evaluating the integral involves finding the antiderivative of the function, which outputs:\(3.0x + 0.25x^2\)
• Substitute the bounds (0 and 4) to find the difference, as in definite integration:\[[3.0x + 0.25x^2]_0^4\]

Here, the integral finds the total work done on the particle by the force along its path, leading to the conclusion that correctly applying integration simplifies the complexity of dealing with non-constant forces.

Kinetic Energy Applications

Kinetic energy, the energy of motion, is closely linked to the concept of work, particularly when analyzing the movement of particles and forces acting upon them. When force does work on a particle, it causes a change in the particle's kinetic energy. This relationship is formalized by the work-energy theorem.The work-energy theorem states:

• The work done on an object equals the change in its kinetic energy.
• In mathematical terms:\[W = \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2)\]where \(m\) is mass, \(v_f\) is final velocity, and \(v_i\) is initial velocity.

In the context of our particle and force example:

• The work calculated, \(16 \, Joules\), represents how much energy is transferred by the force as the particle moves along its path from \(x=0\) to \(x=4.0\, m\).
• Potential alterations in the particle's velocity at these points can be calculated using the energy change.

Understanding the relationship between work and kinetic energy is pivotal. It highlights the practical application of theoretical concepts to real-world scenarios, emphasizing how forces power transitions in the motion of objects.