Question

A force given by \(F(x)=5 x^{3} \hat{x}\left(\right.\) in \(\left.\mathrm{N} / \mathrm{m}^{3}\right)\) 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

Answer: The work done by the force is 1600 J, and the final velocity of the mass is approximately 56.6 m/s.

Step-by-step solution

Calculate the work done by the force

To find the work done by the force, we can use the formula for work as the integral of force with respect to position: \(W = \int_{x_i}^{x_f} F(x) dx\), where \(x_i\) is the initial position and \(x_f\) is the final position. We are given the force function \(F(x) = 5x^3 \hat{x}\) and the initial and final positions: \(x_i = 2.00\,\text{m}\) and \(x_f = 6.00\,\text{m}\). To calculate the work done, we need to evaluate the given integral: \(W = \int_{2}^{6} 5x^3 dx\) Use power rule and solve for W: \(W = 5\left[\frac{1}{4}x^4\right]_{2}^{6}\) \(W = \frac{5}{4}(6^4 - 2^4)\) \(W = \frac{5}{4}(1296 - 16)\) \(W = \frac{5}{4}(1280)\) \(W = 1600\,\text{J}\) The work done by the force is 1600 J.

Apply the work-energy theorem

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy: \(W = \Delta KE\) We know that \(KE = \frac{1}{2}mv^2\), where \(m\) is mass and \(v\) is velocity. The mass of the object in this problem is given as \(m = 1.00\,\text{kg}\). To find the change in kinetic energy, we also need the initial and final velocities. The initial velocity is given as \(v_i = 2.00\,\text{m/s}\). We are tasked to find the final velocity \(v_f\). Using the work-energy theorem: \(W = \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2)\) \(1600\,\text{J} = \frac{1}{2}(1.00\,\text{kg})(v_f^2 - (2.00\,\text{m/s})^2)\)

Solve for the final velocity

Now, we can solve for the final velocity \(v_f\): \(1600\,\text{J} = \frac{1}{2}(1.00\,\text{kg})(v_f^2 - 4.00)\) \(3200\,\text{J} = v_f^2 - 4.00\) \(v_f^2 = 3204\) \(v_f = \sqrt{3204}\) \(v_f ≈ 56.6\,\text{m/s}\) The final velocity of the object is approximately \(56.6\,\text{m/s}\).

Key concepts

Force and Motion

Understanding the connection between force and motion is fundamental to physics, especially when analyzing the behavior of objects under various forces. A force can be thought of as a push or a pull acting on an object, and it is what causes objects to accelerate, decelerate, or change direction. Force is a vector quantity, meaning it has both magnitude and direction, which is why the force in our exercise is described with both a numerical value and the direction \(\hat{x}\).

In the exercise, a force given by \(F(x)=5 x^{3} \hat{x}\) is applied to a mass. This force varies with the cube of the position \(x\), making it a non-constant force; hence, motion analysis in this scenario is more complex than simple linear motion. The use of calculus, namely integration, allows us to determine the work done by the force as the object moves from one position to another.

It's important to keep in mind that when calculating the work done on an object, the direction of the force in relation to the object's displacement is crucial. Since the mass moves along the x-axis from \(x=2.00\,\text{m}\) to \(x=6.00\,\text{m}\), the force does positive work, adding energy to the system and thereby increasing the object's speed.

Kinetic Energy

Kinetic energy (KE) is the energy possessed by an object due to its motion. In classical mechanics, it is calculated with the formula \(KE = \frac{1}{2}mv^2\), where \(m\) represents the mass of the object and \(v\) its velocity. From this formula, you can see that kinetic energy is directly proportional to the mass and the square of the velocity of the object. Thus, if an object's velocity increases (as is the case when work is done on it), its kinetic energy will increase quadratically.

Work-Energy Theorem

In the context of the work-energy theorem, the change in kinetic energy (\(\Delta KE\)) of an object is equal to the net work (\(W\)) done on it. This connects the concepts of work and kinetic energy directly, serving as the cornerstone for solving the exercise's problem b. Using the theorem, we were able to derive the final speed of the mass after work was performed on it, showcasing the transformation of work into kinetic energy, or motion.

Integration in Physics

Integration is a powerful mathematical tool used in physics to determine quantities that are accumulated over intervals. In our exercise, integration plays a key role in determining the amount of work done by a force over a distance. The work done when moving an object with a force that varies with position cannot be found using simple multiplication; this is where integration becomes essential.

In our problem, we were given a force that changes as the object moves along the x-axis. The total work done by this force was found by integrating the force function with respect to position, from the starting point \(x_i\) to the ending point \(x_f\).

• Integration allows us to calculate the total work done across the path, even when the force varies.
• It provides a way to accumulate small amounts of work done over infinitesimally small segments of the path.
• By evaluating the definite integral, we obtain the exact amount of work done on the object.

This result, combined with the work-energy theorem, leads us to determine the final velocity of the object after moving under the influence of the force. Integration thus bridges the gap between abstract force functions and physical outcomes such as motion and energy change in objects.