The force on a particle is directed along an `x` axis and given by `F= F_(0) (x//x_(0)-1)`. Find the work done by the force in moving the particle from `x=0` to `x= 2x_(0)` by (a) plotting `F(x)` and measuring the work from the graph and (b) integrating `F(x)`.

Answer

To solve the problem of finding the work done by the force \( F = F_0 \left( \frac{x}{x_0} - 1 \right) \) as the particle moves from \( x = 0 \) to \( x = 2x_0 \), we will follow two approaches: (a) plotting the graph of \( F(x) \) and measuring the work done from the graph, and (b) integrating \( F(x) \).

### Part (a): Plotting the Graph

1. **Identify the Force Function**: The force function is given as:
\[ F(x) = F_0 \left( \frac{x}{x_0} - 1 \right) \]

2. **Determine Key Points**:
- At \( x = 0 \):
\[ F(0) = F_0 \left( \frac{0}{x_0} - 1 \right) = -F_0 \]
- At \( x = x_0 \):
\[ F(x_0) = F_0 \left( \frac{x_0}{x_0} - 1 \right) = 0 \]
- At \( x = 2x_0 \):
\[ F(2x_0) = F_0 \left( \frac{2x_0}{x_0} - 1 \right) = F_0 \]

3. **Plot the Points**:
- Plot the points \( (0, -F_0) \), \( (x_0, 0) \), and \( (2x_0, F_0) \) on a graph with \( x \) on the horizontal axis and \( F \) on the vertical axis.

4. **Draw the Graph**: Connect the points to form a straight line from \( (0, -F_0) \) to \( (x_0, 0) \) and then from \( (x_0, 0) \) to \( (2x_0, F_0) \).

5. **Calculate the Area Under the Curve**:
- The area under the curve from \( x = 0 \) to \( x = x_0 \) (triangle with base \( x_0 \) and height \( -F_0 \)):
\[ W_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x_0 \times (-F_0) = -\frac{1}{2} F_0 x_0 \]
- The area under the curve from \( x = x_0 \) to \( x = 2x_0 \) (triangle with base \( x_0 \) and height \( F_0 \)):
\[ W_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x_0 \times F_0 = \frac{1}{2} F_0 x_0 \]

6. **Total Work Done**:
\[ W = W_1 + W_2 = -\frac{1}{2} F_0 x_0 + \frac{1}{2} F_0 x_0 = 0 \]

### Part (b): Integrating the Force

1. **Set Up the Integral**:
The work done \( W \) is given by the integral of the force over the displacement:
\[ W = \int_{0}^{2x_0} F(x) \, dx = \int_{0}^{2x_0} F_0 \left( \frac{x}{x_0} - 1 \right) \, dx \]

2. **Integrate**:
\[ W = F_0 \int_{0}^{2x_0} \left( \frac{x}{x_0} - 1 \right) \, dx \]
\[ = F_0 \left[ \frac{1}{x_0} \cdot \frac{x^2}{2} - x \right]_{0}^{2x_0} \]
\[ = F_0 \left[ \frac{1}{x_0} \cdot \frac{(2x_0)^2}{2} - 2x_0 \right] - F_0 \left[ 0 - 0 \right] \]
\[ = F_0 \left[ \frac{2x_0^2}{x_0} - 2x_0 \right] = F_0 \left[ 2x_0 - 2x_0 \right] = 0 \]

### Conclusion
The work done by the force in moving the particle from \( x = 0 \) to \( x = 2x_0 \) is \( 0 \).