Problem 84 - University Physics with Modern Physics

Question

An object has several forces acting on it. One force is \(\overrightarrow{\boldsymbol{F}}=\alpha x \hat{\imath},\) a force in the \(x\) -direction whose magnitude depends on the position of the object. (See Problem \(6.96 . )\) The constant is \(\alpha=2.00 \mathrm{N} / \mathrm{m}^{2} .\) The object moves along the following path: (1) It starts at the origin and moves along the \(y\) -axis to the point \(x=0\) , \(y=1.50 \mathrm{m} ;(2)\) it moves parallel to the \(x\) -axis to the point \(x=1.50 \mathrm{m}, y=1.50 \mathrm{m} ;(3)\) it moves parallel to the \(y\) -axis to the point \(x=1.50 \mathrm{m}, y=0 ;(4)\) it moves parallel to the \(x\) -axis back to the origin. (a) Sketch this path in the \(x y\) -plane. (b) Calculate the work done on the object by \(\overrightarrow{\boldsymbol{F}}\) for each leg of the path and for the complete round trip. (c) Is \(\overrightarrow{\boldsymbol{F}}\) conservative or nonconservative? Explain.

Step-by-Step Solution

Answer

Total work is 0 J; the force \(\overrightarrow{F}\) is conservative.

Step 1: Sketch the Path

To visualize the problem, we draw the path on the \(xy\)-plane based on the points given. Start at the origin \((0,0)\), move vertically up to \((0, 1.5)\), then horizontally to \((1.5, 1.5)\), vertically down to \((1.5, 0)\), and finally horizontally back to the origin \((0,0)\). The path forms a rectangle.

Step 2: Calculate Work on Leg 1

Leg 1 runs from \((0,0)\) to \((0,1.5)\) along the \(y\)-axis where \(x = 0\). Since \(\overrightarrow{F} = \alpha x \hat{\imath}\) acts in the \(x\)-direction, and \(x=0\), the force is zero. Hence, the work done is \[ W_1 = \int_{0}^{1.5} 0 \ dy = 0.\]

Step 3: Calculate Work on Leg 2

Leg 2 moves the object horizontally from \((0, 1.5)\) to \((1.5, 1.5)\). Here, the force \(\overrightarrow{F} = \alpha x \hat{\imath}\) acts in the \(x\)-direction. The work done when moving from \(x=0\) to \(x=1.5\) is \[W_2 = \int_{0}^{1.5} \alpha x \, dx = \frac{\alpha}{2} x^2 \Bigg|_0^{1.5} = \frac{2.00}{2} \times (1.5)^2 = 2.25 \, \text{J}.\]

Step 4: Calculate Work on Leg 3

Leg 3 goes from \((1.5,1.5)\) to \((1.5,0)\) along the \(y\)-axis with \(x = 1.5\) being constant. Hence, the force is constant, and since there's no work done in moving vertically, \[W_3 = \int_{1.5}^{0} \alpha x \, dy = 0.\]

Step 5: Calculate Work on Leg 4

Leg 4 moves horizontally from \((1.5, 0)\) back to \((0, 0)\). The force is directed in the \(x\)-direction opposite to movement, given by \[W_4 = \int_{1.5}^{0} \alpha x \, (-dx) = -\int_{0}^{1.5} \alpha x \, dx = -2.25 \, \text{J}.\]

Step 6: Calculate Total Work

Calculate the total work done over the complete path: \[W_{\text{total}} = W_1 + W_2 + W_3 + W_4 = 0 + 2.25 + 0 - 2.25 = 0 \, \text{J}.\]

Step 7: Determine Conservativeness of the Force

Since the total work done around the closed loop (round trip) is zero, the force \(\overrightarrow{F} = \alpha x \hat{\imath}\) is conservative. In a conservative system, the work done is path-independent.

Key Concepts

Work-Energy Principle

The Work-Energy Principle is a foundational idea in physics that connects the work done by forces to changes in a system's energy. It states that the total work done by all forces acting on an object is equal to the change in its kinetic energy. When we talk about work done by forces, we're looking at how the forces move or change the energy of an object.
In our exercise, we calculate the work done by a force \(\overrightarrow{\boldsymbol{F}}=\alpha x \hat{\imath}\) along different segments of a path. Each segment shows how the force contributes to the overall energy changes. Understanding this principle helps determine if all the work applied to move an object changes its speed or direction.
The Work-Energy Principle becomes especially relevant when dealing with forces that can change or maintain energy within a closed system. For conservative forces, like in this problem, the principle is significant as it shows energy remains constant through reversible paths.

Path Independence

Path independence refers to a situation where the work done by forces does not depend on the path taken but only on the initial and final positions. This is a characteristic feature of conservative forces. If the work done by a force does not vary with paths between two points, the force is path-independent.
In the example exercise, the force \(\overrightarrow{F} = \alpha x \hat{\imath}\) has been shown to have zero total work during a closed loop journey. This indicates that the energy remains unaffected by the particular circumference the object takes, suggesting path independence.
This concept ties into the idea of potential energy, as one can define a potential energy function from which the force can be derived. In conservative systems, potential energy changes only depend on position, not the journey taken between points.

Closed Loop Integration

Closed loop integration involves calculating the work for a force along a closed path. In mathematical terms, this means integrating the force over a round trip, ending at the starting point. For conservative forces, this closed loop integration results in zero work.
For our specific scenario, the path along which the object travels is closed, forming a rectangle. The integration of the force \(\overrightarrow{F} = \alpha x \hat{\imath}\) along this path yields zero net work. This result reinforces the characteristic of conservative forces, where no net energy is added or removed from the system through such a loop.

• Leg 1: No work due to no motion in the force's direction.
• Leg 2: Positive work as the object is aligned with the force.
• Leg 3: Again, no work since movement doesn't align with force.
• Leg 4: Negative work, counteracting work from Leg 2.

Ultimately, adding these works confirms zero total work over the loop, demonstrating that the force is conservative.

Force Dependence on Position

Force dependence on position means that a force's magnitude and sometimes direction vary based on the object's position in space. This is a critical feature of conservative forces, where the force function typically depends on spatial coordinates.
In our example, the given force \(\overrightarrow{F} = \alpha x \hat{\imath}\) varies with the \(x\)-coordinate and is independent of the \(y\)-coordinate. This dependency demonstrates how the force changes its influence as the object moves in the \(x\)-direction.
Understanding position dependence is crucial in calculating forces in real-world phenomena, such as gravitational force or electrostatic forces, where force vectors can vary with distance between objects. In such systems, knowing how force depends on position allows us to derive potential energy functions, providing insight into larger physical landscapes and dynamics of movement.