Problem 70 - University Physics with Modern Physics

Question

CALC A net force along the \(x\) -axis that has \(x\) -component \(F_{x}=-12.0 \mathrm{N}+\left(0.300 \mathrm{N} / \mathrm{m}^{2}\right) x^{2}\) is applied to a 5.00 \(\mathrm{-kg}\) object that is initially at the origin and moving in the \(-x\) -direction with a speed of 6.00 \(\mathrm{m} / \mathrm{s} .\) What is the speed of the object when it reaches the point \(x=5.00 \mathrm{m} ?\)

Step-by-Step Solution

Answer

The speed is approximately 4.12 m/s.

Step 1: Analyze the Force Expression

The force acting along the x-axis is given by \( F_x = -12.0 \, \text{N} + (0.300 \, \text{N/m}^2) x^2 \). This is a function of position \( x \), which means that the force varies as the object moves along the x-axis.

Step 2: Determine Work Done by the Force

The work done by the force when the object moves from an initial position \( x_0 = 0 \) to a final position \( x = 5.00 \, \text{m} \) is given by the definite integral \( W = \int_{0}^{5.00} F_x \, dx \). This integral will help us find the change in kinetic energy.

Step 3: Calculate the Definite Integral

Compute the integral \( W = \int_{0}^{5.00} \left(-12.0 + 0.300x^2\right) \, dx \):\[W = \left[-12.0x + \frac{0.300}{3}x^3\right]_{0}^{5.00} = \left[-12.0(5.00) + 0.1(5.00)^3 \right] - \left[0\right]\]\[W = (-60.0 + 12.5) \, \text{J} = -47.5 \, \text{J}\]This means 47.5 J of work is done on the object.

Step 4: Apply the Work-Energy Theorem

According to the work-energy theorem, the work done by the net force is equal to the change in kinetic energy of the object, \( W = \Delta KE = KE_f - KE_i \).Let \( v_f \) be the final velocity of the object. The initial kinetic energy, \( KE_i = \frac{1}{2}mv_i^2 \) and the final kinetic energy \( KE_f = \frac{1}{2}mv_f^2 \).

Step 5: Solve for the Final Velocity

Substitute the known values and solve for \( v_f \):\[ -47.5 = \frac{1}{2}(5.00)v_f^2 - \frac{1}{2}(5.00)(-6.00)^2 \]\[ -47.5 = \frac{1}{2}(5.00)v_f^2 - 90.0 \]\[ 42.5 = \frac{1}{2}(5.00)v_f^2 \]\[ 85.0 = 5.00v_f^2 \]\[ v_f^2 = 17.0 \]\[ v_f = \sqrt{17.0} \approx 4.12 \, \text{m/s} \]The final speed of the object is approximately \( 4.12 \, \text{m/s} \).

Key Concepts

Net Force

The concept of net force is crucial when analyzing how an object moves under the influence of multiple forces. Net force is essentially the vector sum of all the individual forces acting on an object. It determines whether the object's motion changes, and by how much. In our specific problem, the net force along the x-axis is described by a function of position:

• \( F_x = -12.0 \, \text{N} + (0.300 \, \text{N/m}^2) x^2 \)

This force is not constant; it changes with the object's position along the x-axis. As the object moves, this function accounts for how much force is exerted on the object at each point along its path. This variable force is critical in calculating the work done, as it affects the object's overall energy and ultimately, its velocity. Understanding net force is the key to using the work-energy theorem effectively.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It's given by the expression:

• \( KE = \frac{1}{2}mv^2 \)

where \(m\) is the mass of the object and \(v\) is its velocity. In this problem, kinetic energy plays a pivotal role as we use the work-energy theorem to find the final speed of the object. Initially, the object is moving with a velocity of -6.00 m/s, so its initial kinetic energy can be calculated as:

• \( KE_i = \frac{1}{2}(5.00)(-6.00)^2 \)

As the object moves and work is done on it by the net force, its kinetic energy changes. This change is given by:

• \( \Delta KE = KE_f - KE_i \)

where \( KE_f \) is the final kinetic energy. By calculating the work done and knowing the initial kinetic energy, we can isolate the final velocity, \(v_f\), demonstrating how energy transformations govern motion.

Definite Integral

In physics, a definite integral is often used to calculate the total effect of a varying quantity over a certain range. In this exercise, the definite integral helps us find the total work done by a position-dependent force as the object moves from one point to another:

• \( W = \int_{0}^{5.00} F_x \, dx \)

Calculating this integral with the given force function, \( F_x = -12.0 + 0.300x^2 \), accounts for how the force acts over each small segment of the object's journey. The resulting work obtained from

• \( W = -47.5 \, \text{J} \)

indicates how much energy is transferred to or from the object due to this force over the 5-meter path. Working through definite integrals is a practical application of calculus that enables one to solve real-world physics problems involving forces that change with position.