A force `F=(2+x)` acts on a particle in x-direction where F is in newton and x in meter. Find the work done by this force during a displacement from 1. 0 m to x = 2.0 m.

Explanation

To find the work done by the force \( F = (2 + x) \) during the displacement from \( x = 1.0 \, \text{m} \) to \( x = 2.0 \, \text{m} \), we can follow these steps:

### Step 1: Understand the Work Done Formula
The work done \( W \) by a variable force \( F \) over a displacement \( dx \) is given by the integral:
\[ W = \int_{x_1}^{x_2} F \, dx \]
where \( x_1 \) and \( x_2 \) are the initial and final positions, respectively.

### Step 2: Set Up the Integral
In this case, the force is given as:
\[ F = 2 + x \]
We need to calculate the work done from \( x_1 = 1.0 \, \text{m} \) to \( x_2 = 2.0 \, \text{m} \):
\[ W = \int_{1}^{2} (2 + x) \, dx \]

### Step 3: Break Down the Integral
We can separate the integral into two parts:
\[ W = \int_{1}^{2} 2 \, dx + \int_{1}^{2} x \, dx \]

### Step 4: Calculate Each Integral
1. **First Integral:**
\[ \int_{1}^{2} 2 \, dx = 2 \cdot (x) \bigg|_{1}^{2} = 2 \cdot (2 - 1) = 2 \]

2. **Second Integral:**
\[ \int_{1}^{2} x \, dx = \frac{x^2}{2} \bigg|_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = 2 - 0.5 = 1.5 \]

### Step 5: Combine the Results
Now, we combine the results of the two integrals:
\[ W = 2 + 1.5 = 3.5 \, \text{J} \]

### Conclusion
The work done by the force during the displacement from \( 1.0 \, \text{m} \) to \( 2.0 \, \text{m} \) is:
\[ \boxed{3.5 \, \text{J}} \]