Problem 87 - University Physics with Modern Physics

Question

A variable force acting on a 0.100 - \(\mathrm{kg}\) 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 b) OQS c) OS d) \(O P S Q O\) e) \(O Q S P O\)

Step-by-Step Solution

Answer

The total work done along the path OPS is \(W_{OPS} = \frac{2000}{3} J\). b) What is the total work done along the path OQS? The total work done along the path OQS is \(W_{OQS} = \frac{2000}{3} J\). c) What is the total work done along the straight path OS? The total work done along the straight path OS is \(W_{OS} = \frac{2000}{3} J\). d) What is the total work done along the path OPSQO? The total work done along the path OPSQO is \(W_{OPSQO} = \frac{4000}{3} J\). e) What is the total work done along the path OQSPQ? The total work done along the path OQSPQ is \(W_{OQSPQ} = \frac{4000}{3} J\).

Step 1: a) Path OPS

We can parameterize the path from O (0, 0) to P (0, 10) to S (10, 10) as: 1. From O to P: \(r_1(t) = (0,10t)\); with \(0 \leq t \leq 1\) 2. From P to S: \(r_2(t) = (10t,10)\); with \(0 \leq t \leq 1\) For each path, compute the line integral of the force, then sum them together for the total work done along the path OPS. For O to P, we can write: \(F(r_1(t)) = (0^2 \hat{x} + (10t)^2 \hat{y}) = (0, 100t^2 \hat{y})\) \(r_1'(t) = (0, 10)\) \(F \cdot r_1'(t) = (0, 100t^2) \cdot (0, 10) = 1000t^2\) \(W_{OP} = \int_{0}^{1} 1000t^2 dt = \frac{1000}{3}t^3\big|_{0}^1 = \frac{1000}{3} J\) For P to S, we can write: \(F(r_2(t)) = (10t^2 \hat{x} + 10^2 \cdot \hat{y}) = (100t^2 \hat{x}, 100 \cdot \hat{y})\) \(r_2'(t) = (10, 0)\) \(F \cdot r_2'(t) = (100t^2, 100) \cdot (10, 0) = 1000t^2\) \(W_{PS} = \int_{0}^{1} 1000t^2 dt = \frac{1000}{3}t^3\big|_{0}^1 = \frac{1000}{3} J\) The total work done along the path OPS is: \(W_{OPS} = W_{OP} + W_{PS} = \frac{1000}{3} + \frac{1000}{3} = \frac{2000}{3} J\)

Step 2: b) Path OQS

We can parameterize the path from O (0, 0) to Q (10, 0) to S (10, 10) as: 1. From O to Q: \(r_3(t) = (10t,0)\); with \(0 \leq t \leq 1\) 2. From Q to S: \(r_4(t) = (10,10t)\); with \(0 \leq t \leq 1\) For each path, compute the line integral of the force, then sum them together for the total work done along the path OQS. For O to Q, we can write: \(F(r_3(t)) = ((10t)^2 \hat{x},0) = (100t^2,0)\) \(r_3'(t) = (10,0)\) \(F \cdot r_3'(t) = (100t^2,0)\cdot(10,0) = 1000t^2\) \(W_{OQ} = \int_{0}^{1} 1000t^2 dt = \frac{1000}{3}t^3\big|_{0}^1 = \frac{1000}{3} J\) For Q to S, we can write: \(F(r_4(t)) = (10^2,(10t)^2) = (100,100t^2)\) \(r_4'(t) = (0,10)\) \(F \cdot r_4'(t) = (100,100t^2)\cdot(0,10) = 1000t^2\) \(W_{QS} = \int_{0}^{1} 1000t^2 dt = \frac{1000}{3}t^3\big|_{0}^1 = \frac{1000}{3} J\) The total work done along the path OQS is: \(W_{OQS} = W_{OQ} + W_{QS} = \frac{1000}{3} + \frac{1000}{3} = \frac{2000}{3} J\)

Step 3: c) Path OS

We can parameterize the path from O (0, 0) to S (10, 10) along a straight line as: 1. From O to S: \(r_5(t) = (10t, 10t)\); with \(0 \leq t \leq 1\) Compute the line integral of the force along the straight path OS. \(F(r_5(t)) = ((10t)^2, (10t)^2) = (100t^2, 100t^2)\) \(r_5'(t) = (10, 10)\) \(F \cdot r_5'(t) = (100t^2, 100t^2) \cdot (10, 10) = 1000t^2 + 1000t^2 = 2000t^2\) \(W_{OS} = \int_{0}^{1} 2000t^2 dt = \frac{2000}{3}t^3\big|_{0}^1 = \frac{2000}{3} J\)

Step 4: d) Path OPSQO

For the path \(O P S Q O\), we can use the results from (a) and (b), considering that the work done for the path \(QO\) equals to the work done for the path \(OQ\): \(W_{OPSQO} = W_{OP} + W_{PS} + W_{QS} + W_{OQ} = \frac{1000}{3} + \frac{1000}{3} + \frac{1000}{3} + \frac{1000}{3} = \frac{4000}{3} J\)

Step 5: e) Path OQSPQ

For the path \(O Q S P O\), we can also use the results from (a) and (b), and consider that the work done for the path \(PS\) equals to the work done for the path \(SP\): \(W_{OQSPQ} = W_{OQ} + W_{QS} + W_{PS} + W_{OP} = \frac{1000}{3} + \frac{1000}{3} + \frac{1000}{3} + \frac{1000}{3} = \frac{4000}{3} J\)

Key Concepts

Variable Force

In physics, a variable force is a force that changes in magnitude and/or direction as an object moves. In our exercise, the force acting on a particle in the xy-plane is expressed as a function of the particle's position in space, represented by the coordinates (x, y). The given force is F(x, y) = (x^2 \(\hat{x}\) + y^2 \(\hat{y}\)), meaning that the force components vary with the square of their respective position coordinates.

This is significant since the work done by such forces cannot be calculated using the simple formula \( W = F \times d \times \cos(\theta) \) applicable to constant forces. Instead, the concept of a line integral must be used, integrating the force over the path taken by the particle. This requires a deeper understanding of calculus, particularly when the path is not straight or the force is not constant in direction or magnitude.

Work Done by Force

The work done by a force is a measure of energy transferred by the force when an object moves a certain distance. The work along a path in a varying force field is found using a line integral, which is defined as the integral of the force dotted with the differential element of the path.

In our problem, we calculate the work done over different paths using the line integral method, symbolically shown as \( W = \int_{C} \mathbf{F} \cdot d\mathbf{r} \), where \( \mathbf{F} \) is the force vector, \( d\mathbf{r} \) is the differential element along the path, and \( C \) denotes the specific path taken. When evaluating this integral, we account for the fact that the work done is path-dependent, which is a key characteristic when dealing with variable forces.

Parameterization of Path

The parameterization of a path involves describing a path as a set of equations that define the location of a point along the path at any given time or position parameter. In the given problem, separate paths are parameterized for different segments of the particle's journey. For example, the path from point O to P is expressed as \( r_1(t) = (0, 10t) \), with t varying between 0 and 1.

Such parameterizations are crucial as they convert the line integral into a definite integral over the parameter t, making it possible to calculate the work done by the force. To correctly calculate the work, we first need to compute the derivative of the parameterized path, and then evaluate the integrals along each segmented path. Effective parameterization simplifies complex paths into manageable equations that yield to the process of integration, allowing us to determine the work for any given path.