Question

A variable force acting on a 0.100 -kg particle moving in the \(x y\) -plane is given by \(F(x, y)=\left(x^{2} \hat{x}+y^{2} \hat{y}\right) \mathrm{N},\) where \(x\) and \(y\) are in meters. Suppose that due to this force, the particle moves from the origin, \(O\), to point \(S\), with coordinates \((10.0 \mathrm{~m}, 10.0 \mathrm{~m})\). The coordinates of points \(P\) and \(Q\) are \((0 \mathrm{~m}, 10.0 \mathrm{~m})\) and \((10.0 \mathrm{~m}, 0 \mathrm{~m})\), respectively. Determine the work performed by the force as the particle moves along each of the following paths: a) OPS c) \(O S\) e) \(O Q S P O\) b) \(O Q S\) d) \(O P S Q O\)

Short answer

Based on the analysis and solution, calculate the total work done by the force F(x, y) = (x^2 * x-hat + y^2 * y-hat) N as the particle moves along each given path. a) Path OPS: The total work done = 110000/3 J b) Path OQS: The total work done = 110000/3 J c) Path OS: The total work done = 200000/3 J d) Path OPSQO: The total work done = 110000/3 J e) Path OQSPO: The total work done = 110000/3 J

Step-by-step solution

Set up the Integral

Since the path consists of two straight-line segments (OP and PS), we will calculate the work done along each segment separately and then add them up. First, we need to parameterize each segment in terms of a single parameter, typically t. For OP: \(r_{\text{OP}}(t)=\langle 0, 10t \rangle, \; 0 \leq t\leq 1\) For PS: \(r_{\text{PS}}(t)=\langle 10t, 10 \rangle, \; 0 \leq t\leq 1\)

Calculate Line Integrals

Now, we need to calculate the integral of \(F(r) \cdot dr\) for each segment and add them up. For OP: \(\int_{0}^{1} F(r_{\text{OP}}(t)) \cdot r_{\text{OP}}'(t) \, dt\) For PS: \(\int_{0}^{1} F(r_{\text{PS}}(t)) \cdot r_{\text{PS}}'(t) \, dt\)

Evaluate Line Integrals

After evaluating the line integrals, we can find the work done along each segment: For OP: \(W_{\text{OP}} = \int_{0}^{1} (x^2\hat{x}+y^2\hat{y}) \cdot (0\hat{x}+10\hat{y}) \, dt = \int_{0}^{1} 10000t^2 \, dt = \frac{10000}{3} (1-0)^3= \frac{10000}{3} \,\text{J}\) For PS: \(W_{\text{PS}} = \int_{0}^{1} (10000t^2\hat{x}+100\hat{y}) \cdot (10\hat{x}+0\hat{y}) \, dt = \int_{0}^{1} 100000t^2 \, dt = \frac{100000}{3} (1-0)^3 = \frac{100000}{3} \,\text{J}\) Finally, we add both results to get the work done along path OPS: \(W_{OPS}=W_{OP}+W_{PS}=\frac{10000}{3}+\frac{100000}{3}=\frac{110000}{3}\;\text{J}\) #c) Path: OS#

Set up the Integral

We need to evaluate the work done as the particle moves along the straight line from O to S. First, we will parameterize the path: For OS: \(r_{\text{OS}}(t)=\langle 10t, 10t \rangle, \; 0 \le t\le 1\)

Calculate the Line Integral

Now we will calculate the integral of \(F(r) \cdot dr\) for the path OS: \(W_{\text{OS}} = \int_{0}^{1} F(r_{\text{OS}}(t)) \cdot r_{\text{OS}}'(t) \, dt\)

Evaluate the Line Integral

After evaluating the line integral, we can find the work done: \(W_{\text{OS}} = \int_{0}^{1} (100x^2\hat{x}+100y^2\hat{y}) \cdot (10\hat{x}+10\hat{y}) \, dt = \int_{0}^{1} 1000(10t)^2+1000(10t)^2 \, dt = \int_{0}^{1} 200000t^2 \, dt = \frac{200000}{3}(1-0)^3 = \frac{200000}{3} \;\text{J}\) To save space, we only show the results for the remaining paths: #e) Path: OQSPO# \(W_{OQ} = \frac{100000}{3} \;\text{J}\), \(W_{QS} = \frac{10000}{3} \;\text{J}\), \(W_{SP} = 0 \;\text{J}\), \(W_{PO} = 0 \;\text{J}\) Total work \(W_{OQSP}=\frac{110000}{3} \;\text{J}\). #b) Path: OQS# \(W_{OQ} = \frac{100000}{3} \;\text{J}\), \(W_{QS} = \frac{10000}{3} \;\text{J}\) Total work \(W_{OQS}=\frac{110000}{3} \;\text{J}\) #d) Path: OPSQO# \(W_{OP} = \frac{10000}{3} \;\text{J}\), \(W_{PS} = \frac{100000}{3} \;\text{J}\), \(W_{SQ} = 0 \;\text{J}\), \(W_{QO} = 0 \;\text{J}\) Total work \(W_{OPSQO}=\frac{110000}{3} \;\text{J}\)

Key concepts

Line Integral in Physics

In physics, the line integral is a crucial concept, especially when calculating work, electric potential, or magnetic flux. It helps us understand how different quantities add up along a specific path in space. To illustrate, imagine a particle is being moved along a path by a force. The line integral essentially sums up the work done by the force at every infinitesimal segment of the path.

For a force vector field, represented by a function of coordinates, and a path parameterized by a variable, the work done is given by the integral of the dot product of the force vector and the differential element of the path. It encapsulates the notion that work is a scalar product: force times the displacement in the direction of the force. This concept was crucial in evaluating the work done as the particle moved from the origin O to point S along different paths in our textbook problem.

Force Vector Field

A force vector field is a map that assigns to every point in space a force vector. This vector represents the force experienced by a particle located at that point. In our problem, the force vector field is given by F(x, y) = \(x^2 \hat{x} + y^2 \hat{y}\), where \(x^2 \hat{x}\) and \(y^2 \hat{y}\) are the components of the force in the x and y directions, respectively.

Such a force field can be visualized as arrows of varying lengths and directions, representing the force at different points. For instance, near the origin, the force vectors are small since both x and y are small, but they grow larger as one moves away from the origin. Exploring the force field helps in conceptualizing the problem and understanding the variable force acting on the moving particle.

Parameterization of Path

Parameterizing the path provides a means to describe a curve, or a move from one point to another, with a single variable, often denoted as t. In the context of work and energy, this method allows us to express the position and the force as functions of this variable. Consequently, we can calculate the work done over a path as t varies, usually within a specified range.

In our exercise, we parameterized segments OP and PS to calculate the work done along these paths. For instance, for segment OP, \(r_{\text{OP}}(t)=\langle 0, 10t \rangle\) where t varied from 0 to 1. Such parameterizations transform the work done calculation into a more manageable form and simplify the process of evaluating the line integrals.

Work-Energy Theorem

The work-energy theorem is a foundational principle in physics that states the work done by all forces acting on a particle causes a change in the kinetic energy of the particle. This theorem is powerful because it connects two seemingly distinct concepts: work (a force applied over a distance) and energy (the capacity to perform work).

In practical terms, if a particle is acted upon by a force along a displacement, the work done on the particle changes its kinetic energy. This theorem reinforces our approach to the exercise, where the work done on a particle moving along various paths in a force field was computed. Understanding the theorem helps students relate the physical movement of the particle and the resulting energetic changes, giving deeper insight into physical processes.