Math Problem Statement

Calculate the work done along a path y = 0.6x^2 from (0, 0) to (2, 2.4), given the force F = 3yhat{i} + 7yhat{j}.

Solution

Let's solve the problem step by step while addressing all the requested elements.

Given:

  1. Path: y=0.6x2y = 0.6x^2, from (0,0)(0, 0) to (2,2.4)(2, 2.4).
  2. Force: F=3yi^+7yj^\vec{F} = 3y\hat{i} + 7y\hat{j}.

We need to calculate the work done along this path.


a. Diagram depicting the trajectory of motion

The path is a parabola defined by y=0.6x2y = 0.6x^2, and the object travels from (0,0)(0, 0) to (2,2.4)(2, 2.4).

(I will sketch this trajectory later if you need a visual representation.)


b. Equation showing the infinitesimal element of work

The infinitesimal work done dWdW is given by: dW=FdrdW = \vec{F} \cdot d\vec{r} where dr=dxi^+dyj^d\vec{r} = dx\hat{i} + dy\hat{j} is the infinitesimal displacement vector.

Substitute F=3yi^+7yj^\vec{F} = 3y\hat{i} + 7y\hat{j} and drd\vec{r}: dW=(3yi^+7yj^)(dxi^+dyj^)dW = (3y\hat{i} + 7y\hat{j}) \cdot (dx\hat{i} + dy\hat{j}) dW=3ydx+7ydydW = 3y \, dx + 7y \, dy


c. Mathematical working to find the total work

  1. Express dydy in terms of xx:

    Since y=0.6x2y = 0.6x^2, differentiate to find dydy: dy=ddx(0.6x2)dx=1.2xdxdy = \frac{d}{dx}(0.6x^2) \, dx = 1.2x \, dx

  2. Substitute y=0.6x2y = 0.6x^2 and dy=1.2xdxdy = 1.2x \, dx into dWdW:

    dW=3ydx+7ydydW = 3y \, dx + 7y \, dy dW=3(0.6x2)dx+7(0.6x2)(1.2xdx)dW = 3(0.6x^2) \, dx + 7(0.6x^2)(1.2x \, dx) Simplify: dW=1.8x2dx+5.04x3dxdW = 1.8x^2 \, dx + 5.04x^3 \, dx dW=(1.8x2+5.04x3)dxdW = (1.8x^2 + 5.04x^3) \, dx

  3. Integrate dWdW from x=0x = 0 to x=2x = 2: W=02(1.8x2+5.04x3)dxW = \int_0^2 (1.8x^2 + 5.04x^3) \, dx

    Split into two integrals: W=021.8x2dx+025.04x3dxW = \int_0^2 1.8x^2 \, dx + \int_0^2 5.04x^3 \, dx

    Calculate each term: 021.8x2dx=1.8x3302=1.883=4.8\int_0^2 1.8x^2 \, dx = 1.8 \cdot \frac{x^3}{3} \Big|_0^2 = 1.8 \cdot \frac{8}{3} = 4.8 025.04x3dx=5.04x4402=5.04164=20.16\int_0^2 5.04x^3 \, dx = 5.04 \cdot \frac{x^4}{4} \Big|_0^2 = 5.04 \cdot \frac{16}{4} = 20.16

    Add the results: W=4.8+20.16=24.96W = 4.8 + 20.16 = 24.96


Final Answer:

The total work done is: 24.96J\boxed{24.96 \, \text{J}}

Would you like a visual diagram, detailed explanation, or assistance with any part of this?


Follow-up Questions:

  1. How is the concept of work in this problem related to the dot product of vectors?
  2. What would happen if the force field F\vec{F} varied with xx instead of yy?
  3. Why is it important to parametrize the path in terms of xx?
  4. What is the physical meaning of the work done here?
  5. Could the work be solved using a numerical method if integration was too complex?

Tip:

When dealing with path integrals, always ensure the force and displacement components align correctly with the curve's parametrization.

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Math Problem Analysis

Mathematical Concepts

Path Integrals
Work and Energy
Dot Product
Differential Calculus
Integration

Formulas

dW = F · dr
dW = (3y dx + 7y dy)
dy = 1.2x dx
W = ∫(1.8x² + 5.04x³) dx

Theorems

Work-Energy Theorem
Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate (Physics/Engineering)