A force `F=(10+0.50x)` acts on a particle in the x direction, where F is in newton and x in meter. Find the work done by this force during a displacement form x=0 to x=2.0m

Options

A. 31.5 J
B. 63 J
C. 21 J
D. 42 J

Explanation

To find the work done by the force \( F = (10 + 0.50x) \) during the displacement from \( x = 0 \) 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 can be calculated using the integral of the force over the displacement. Mathematically, this is expressed as:
\[ W = \int_{x_1}^{x_2} F \, dx \]
where \( F \) is the force as a function of \( x \), and \( x_1 \) and \( x_2 \) are the initial and final positions.

### Step 2: Set Up the Integral
Given the force \( F = 10 + 0.50x \), we need to set up the integral from \( x = 0 \) to \( x = 2 \):
\[ W = \int_{0}^{2} (10 + 0.50x) \, dx \]

### Step 3: Break Down the Integral
We can break down the integral into two parts:
\[ W = \int_{0}^{2} 10 \, dx + \int_{0}^{2} 0.50x \, dx \]

### Step 4: Calculate Each Integral
1. **First Integral**:
\[ \int_{0}^{2} 10 \, dx = 10x \bigg|_{0}^{2} = 10(2) - 10(0) = 20 \]

2. **Second Integral**:
\[ \int_{0}^{2} 0.50x \, dx = 0.50 \cdot \frac{x^2}{2} \bigg|_{0}^{2} = 0.50 \cdot \frac{(2)^2}{2} - 0.50 \cdot \frac{(0)^2}{2} = 0.50 \cdot \frac{4}{2} = 0.50 \cdot 2 = 1 \]

### Step 5: Combine the Results
Now, we add the results of the two integrals:
\[ W = 20 + 1 = 21 \, \text{J} \]

### Conclusion
The work done by the force during the displacement from \( x = 0 \) to \( x = 2.0 \, \text{m} \) is:
\[ \boxed{21 \, \text{J}} \]