A force `vecF=(3x^(2)+2x-7)N` acts on a 2 kg body as a result of which the body gets displaced from x=0 to x=5m. The work done by the force will be –

Options

A. 35 Joule
B. 70 Joule
C. 115 Joule
D. 270 Joule

Answer

To find the work done by the force \( \vec{F} = (3x^2 + 2x - 7) \, \text{N} \) on a body as it moves from \( x = 0 \) to \( x = 5 \, \text{m} \), we will use the formula for work done by a variable force:

\[ W = \int_{x_1}^{x_2} F(x) \, dx \]

### Step-by-Step Solution:

1. **Identify the Force Function**:
The force acting on the body is given by:
\[ F(x) = 3x^2 + 2x - 7 \]

2. **Set Up the Integral for Work Done**:
The work done by the force as the body moves from \( x = 0 \) to \( x = 5 \) is given by:
\[ W = \int_{0}^{5} (3x^2 + 2x - 7) \, dx \]

3. **Integrate the Force Function**:
We will integrate the function \( 3x^2 + 2x - 7 \):
\[ W = \int (3x^2) \, dx + \int (2x) \, dx - \int (7) \, dx \]
- The integral of \( 3x^2 \) is \( x^3 \).
- The integral of \( 2x \) is \( x^2 \).
- The integral of \( 7 \) is \( 7x \).

Thus, we have:
\[ W = \left[ x^3 + x^2 - 7x \right]_{0}^{5} \]

4. **Evaluate the Integral at the Limits**:
Now we will evaluate this expression at the upper limit \( x = 5 \) and lower limit \( x = 0 \):
\[ W = \left( 5^3 + 5^2 - 7 \cdot 5 \right) - \left( 0^3 + 0^2 - 7 \cdot 0 \right) \]
- Calculate \( 5^3 = 125 \)
- Calculate \( 5^2 = 25 \)
- Calculate \( 7 \cdot 5 = 35 \)

Therefore:
\[ W = (125 + 25 - 35) - 0 \]
\[ W = 125 + 25 - 35 = 115 \, \text{J} \]

5. **Conclusion**:
The work done by the force as the body moves from \( x = 0 \) to \( x = 5 \) meters is:
\[ W = 115 \, \text{J} \]