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Ch 10: Interactions and Potential Energy
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 10: Interactions and Potential EnergyProblem 65a
Chapter 10, Problem 65a

CALC An object moving in the xy-plane is subjected to the force F⃗=(2xyi^+x2j^)N\(\vec{F}\) = (2xy\,\(\hat{i}\) + x^2\,\(\hat{j}\))\,\(\text{N}\), where x and y are in m. The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis. How much work does the force do?

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Step 1: Recall the formula for work done by a force: \( W = \int \mathbf{F} \cdot d\mathbf{r} \), where \( \mathbf{F} \) is the force vector and \( d\mathbf{r} \) is the infinitesimal displacement vector. In this case, \( \mathbf{F} = (2xy \hat{i} + x^2 \hat{j}) \) and the path is piecewise (first along the x-axis, then along the y-axis).
Step 2: For the first segment of the path (from \( (0, 0) \) to \( (a, 0) \) along the x-axis), the displacement vector is \( d\mathbf{r} = dx \hat{i} \). Substitute \( y = 0 \) into \( \mathbf{F} \), which simplifies to \( \mathbf{F} = x^2 \hat{j} \). Since \( \mathbf{F} \cdot d\mathbf{r} = 0 \) (no \( \hat{i} \) component in \( \mathbf{F} \)), the work done along this segment is zero.
Step 3: For the second segment of the path (from \( (a, 0) \) to \( (a, b) \) along the y-axis), the displacement vector is \( d\mathbf{r} = dy \hat{j} \). Here, \( x = a \) is constant, so \( \mathbf{F} = (2ay \hat{i} + a^2 \hat{j}) \). The dot product \( \mathbf{F} \cdot d\mathbf{r} = a^2 dy \).
Step 4: Integrate \( \mathbf{F} \cdot d\mathbf{r} \) over the second segment of the path. The limits of integration for \( y \) are from \( 0 \) to \( b \). Thus, \( W = \int_0^b a^2 dy \). Since \( a^2 \) is constant, the integral simplifies to \( W = a^2 \int_0^b dy = a^2 [y]_0^b \).
Step 5: Evaluate the integral result. Substitute the limits of integration into \( [y]_0^b \), which gives \( W = a^2 (b - 0) = a^2 b \). This is the total work done by the force along the given path.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Work Done by a Force

Work is defined as the integral of force along a path. In this case, the work done by the force F on the object as it moves from one point to another can be calculated by integrating the force vector along the specified path. The work done is dependent on both the magnitude of the force and the displacement in the direction of the force.
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Line Integral

A line integral is used to calculate the work done by a force field along a specific path. It involves integrating the force vector along the trajectory of the object, which can be broken down into segments if the path is not straight. In this problem, the path consists of two segments: first along the x-axis and then parallel to the y-axis.
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Vector Fields

A vector field assigns a vector to every point in space, representing quantities like force or velocity. In this question, the force F is a vector field defined in the xy-plane, where its components depend on the coordinates x and y. Understanding how to evaluate the force at different points along the path is crucial for calculating the work done.
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Related Practice
Textbook Question

In a physics lab experiment, a compressed spring launches a 20 g metal ball at a 30° angle. Compressing the spring 20 cm causes the ball to hit the floor 1.5 m below the point at which it leaves the spring after traveling 5.0 m horizontally. What is the spring constant?

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Textbook Question

CALC The potential energy for a particle that can move along the x-axis is U = Ax2 + B sin(πx/L), where A, B, and L are constants. What is the force on the particle at (a) x = 0, (b) x = L/2, and (c) x = L?

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Textbook Question

A 10 kg box slides 4.0 m down the frictionless ramp shown in FIGURE CP10.73, then collides with a spring whose spring constant is 250 N/m. What is the maximum compression of the spring?

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Textbook Question

CALC An object moving in the xy-plane is subjected to the force F⃗=(2xyi^+x2j^)N\(\vec{F}\) = (2xy\,\(\hat{i}\) + x^2\,\(\hat{j}\))\,\(\text{N}\), where x and y are in m. Is this a conservative force?

Textbook Question

A particle that can move along the x-axis is part of a system with potential energy U(x) = A/x2 − B/x where A and B are positive constants. Where are the particle's equilibrium positions?

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Textbook Question

A clever engineer designs a 'sprong' that obeys the force law Fx=−q(x−xeq)³ , where xeq is the equilibrium position of the end of the sprong and q is the sprong constant. For simplicity, we'll let xeq = 0 m .Then Fx = −qx³. Find an expression for the potential energy of a stretched or compressed sprong.

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