Problem 68 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

The speed of the object at \(x = 5.00\) m is approximately 4.12 m/s.

Step 1: Set up the Work-Energy Theorem

The work-energy theorem states that the work done by all the forces acting on an object equals the change in its kinetic energy. Mathematically, \( W = \Delta K \), where \( W \) is the work done, and \( \Delta K \) is the change in kinetic energy, given by \( \frac{1}{2} m (v_f^2 - v_i^2) \). Here, \( m = 5.00 \) kg, and we need to find the final velocity \( v_f \) when the object reaches \( x = 5.00 \) m.

Step 2: Calculate Initial Kinetic Energy

Initially, the object moves at a speed of 6.00 m/s in the \(-x\)-direction. The initial kinetic energy is given by:\[K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} (5.00 \, \text{kg}) (6.00 \, \text{m/s})^2 = 90.0 \, \text{J}.\]

Step 3: Calculate Work Done by the Force

The work done by the force as the object moves from \( x = 0 \) to \( x = 5.00 \) m is found by integrating the force over distance:\[W = \int_{0}^{5.00} \left(-12.0 \text{ N} + (0.300 \text{ N/m}^2) x^2\right) dx.\]Integrate term by term:\[W = \left[-12.0x + 0.100x^3\right]_{0}^{5.00}.\]Evaluate this expression:\[W = \left(-12.0 \times 5.00 + 0.100 \times (5.00)^3\right) - \left(-12.0 \times 0 + 0.100 \times 0^3\right) = -60.0 + 12.5 = -47.5 \, \text{J}.\]

Step 4: Calculate Final Kinetic Energy and Speed

Using the work-energy theorem, equate the work done to the change in kinetic energy:\[-47.5 = \frac{1}{2} m (v_f^2 - v_i^2).\]Plug in the known values:\[-47.5 = \frac{1}{2} (5.00) (v_f^2 - 36.0).\]Solving for \( v_f^2 \), we find:\[-47.5 = 2.5 (v_f^2 - 36),\]\[-47.5 = 2.5v_f^2 - 90,\]\[2.5v_f^2 = 42.5,\]\[v_f^2 = 17,\]\[v_f = \sqrt{17} \approx 4.12 \, \text{m/s}.\]

Key Concepts

Kinetic Energy

Kinetic energy is the energy of an object due to its motion. It is a crucial concept in physics because it allows us to understand how objects move and interact with forces. The formula for kinetic energy (K) is:
\[K = \frac{1}{2} m v^2\]
where:

• \(m\) is the mass of the object
• \(v\) is the velocity of the object

In the given problem, the object initially moves with a velocity of -6.00 m/s. Using the formula, we can calculate its initial kinetic energy:
\[K_i = \frac{1}{2} \times 5.00 \, \text{kg} \times (6.00 \, \text{m/s})^2 = 90.0 \, \text{J}\]
The negative sign indicates that the object is moving in the opposite direction of the positive reference axis. However, kinetic energy is a scalar quantity, so it is always positive, regardless of direction. As the object moves, forces can do work on it, changing its kinetic energy and thereby altering its speed. Understanding how kinetic energy changes in response to forces is essential for solving problems involving motion.

Net Force

The net force is the total force acting on an object when all individual forces are considered. It determines the acceleration and direction of motion of the object according to Newton's second law:
\[F_{\text{net}} = m a\]
In the exercise, the net force along the x-axis is given by its x-component:
\[F_x = -12.0 \, \text{N} + 0.300 \, \text{N/m}^2 \times x^2\]
This force has two parts:

• A constant force of -12.0 N that acts constantly as the object moves.
• A variable force that depends on the square of the position \(x^2\) and increases as the object moves away from the origin.

The forces work together to alter the motion of the object. The negative sign in the constant force suggests that it acts in the opposite direction of the object's initial motion. Calculating the net force helps determine how the velocity and kinetic energy of the object change as it moves.

Integration in Physics

Integration is a mathematical technique used in physics to find terms related to area, volume, and other accumulated quantities. When an object moves in response to a force, integration helps in calculating the work done by that force over a distance. In this problem, the work done by the force is calculated using the integral of the force expression over the range from \(x = 0\) to \(x = 5.00\, \text{m}\):
\[W = \int_{0}^{5.00} \left(-12.0 \, \text{N} + 0.300 \, \text{N/m}^2 \, x^2 \right) dx\]
Integration allows us to add up all the tiny values of work done along every small movement from \(x\). For our specific force equation, performing the integration term by term results in:
\[W = \left[-12.0x + 0.100x^3\right]_{0}^{5.00}\]
This process helps identify the total work done as -47.5 J, which represents the change in energy needed to compute the object's final velocity using the work-energy theorem.
Understanding integration is vital for analyzing the effects of varying forces over a path, leading to insights into energy changes and mechanics dynamics.