A position dependent force `F=7-2x+3x^(2)` acts on a small body of mass 2 kg and displaced it from `x=0` to `x=5m`. Calculate the work done in joule.

Options

A. 135
B. 270
C. 35
D. 70

Answer

To calculate the work done by the position-dependent force \( F = 7 - 2x + 3x^2 \) as it acts on a body of mass 2 kg displaced from \( x = 0 \) to \( x = 5 \) m, we will follow these steps:

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

### Step 2: Set Up the Integral
Given the force \( F = 7 - 2x + 3x^2 \), we need to calculate the work done as the body moves from \( x = 0 \) to \( x = 5 \):
\[ W = \int_{0}^{5} (7 - 2x + 3x^2) \, dx \]

### Step 3: Break Down the Integral
We can separate the integral into three parts:
\[ W = \int_{0}^{5} 7 \, dx - \int_{0}^{5} 2x \, dx + \int_{0}^{5} 3x^2 \, dx \]

### Step 4: Calculate Each Integral
1. **First Integral**:
\[ \int_{0}^{5} 7 \, dx = 7x \bigg|_{0}^{5} = 7(5) - 7(0) = 35 \]

2. **Second Integral**:
\[ \int_{0}^{5} 2x \, dx = 2 \cdot \frac{x^2}{2} \bigg|_{0}^{5} = x^2 \bigg|_{0}^{5} = 5^2 - 0^2 = 25 \]

3. **Third Integral**:
\[ \int_{0}^{5} 3x^2 \, dx = 3 \cdot \frac{x^3}{3} \bigg|_{0}^{5} = x^3 \bigg|_{0}^{5} = 5^3 - 0^3 = 125 \]

### Step 5: Combine the Results
Now, substituting back into the expression for work done:
\[ W = 35 - 25 + 125 \]
\[ W = 135 \, \text{joules} \]

### Final Answer
The work done by the force as the body is displaced from \( x = 0 \) to \( x = 5 \) m is \( 135 \, \text{joules} \).

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