A force of `F=(0.5x+12)N` acts on a particle. If x is in metre, calculate the work done by the force during the displacement of the particle from x = 0 to x = 4 m

Options

A. 128 J
B. 80 J
C. 64 J
D. 52 J

Answer

To calculate the work done by the force \( F = (0.5x + 12) \, \text{N} \) during the displacement of the particle from \( x = 0 \) to \( x = 4 \, \text{m} \), we will follow these steps:

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

### Step 2: Set Up the Integral
In our case, the force \( F \) is given as \( F = 0.5x + 12 \). We need to evaluate the integral from \( x = 0 \) to \( x = 4 \):
\[ W = \int_{0}^{4} (0.5x + 12) \, dx \]

### Step 3: Calculate the Integral
We can split the integral into two parts:
\[ W = \int_{0}^{4} 0.5x \, dx + \int_{0}^{4} 12 \, dx \]

Now, we calculate each integral separately.

1. **First Integral**:
\[ \int 0.5x \, dx = 0.5 \cdot \frac{x^2}{2} = 0.25x^2 \]
Evaluating from 0 to 4:
\[ 0.25(4^2) - 0.25(0^2) = 0.25(16) - 0 = 4 \]

2. **Second Integral**:
\[ \int 12 \, dx = 12x \]
Evaluating from 0 to 4:
\[ 12(4) - 12(0) = 48 - 0 = 48 \]

### Step 4: Combine the Results
Now, we add the results of the two integrals:
\[ W = 4 + 48 = 52 \, \text{J} \]

### Final Answer
The work done by the force during the displacement from \( x = 0 \) to \( x = 4 \, \text{m} \) is:
\[ \boxed{52 \, \text{J}} \]