A force `f =10 + 2 x` acts on a particle moving in straight line on x axis the work done by this force during a displacement from `x =0` and `x =3m`

Options

A. `13`J
B. `29`J
C. `39`J
D. `43`J

Answer

To find the work done by the force \( f = 10 + 2x \) during the displacement from \( x = 0 \) to \( x = 3 \) meters, we will 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:
\[ W = \int_{x_1}^{x_2} f \, dx \]
where \( f \) is the force and \( x_1 \) and \( x_2 \) are the initial and final positions.

### Step 2: Set Up the Integral
In this case, the force \( f \) is given by:
\[ f = 10 + 2x \]
We need to integrate this force from \( x = 0 \) to \( x = 3 \):
\[ W = \int_{0}^{3} (10 + 2x) \, dx \]

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

### Step 4: Calculate Each Integral
1. **First Integral:**
\[ \int_{0}^{3} 10 \, dx = 10 \cdot [x]_{0}^{3} = 10 \cdot (3 - 0) = 30 \]

2. **Second Integral:**
\[ \int_{0}^{3} 2x \, dx = 2 \cdot \left[\frac{x^2}{2}\right]_{0}^{3} = [x^2]_{0}^{3} = 3^2 - 0^2 = 9 \]

### Step 5: Combine the Results
Now, we can add the results of the two integrals:
\[ W = 30 + 9 = 39 \, \text{Joules} \]

### Final Answer
The work done by the force during the displacement from \( x = 0 \) to \( x = 3 \) meters is:
\[ \boxed{39 \, \text{Joules}} \]

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