A force`vecF=(7-2x+3x^(2))` N is applied on a 2 kg mass which displaces it from x = 0 to x = 5 m. Work done in joule is -

Options

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

Answer

To find the work done by the force \( \vec{F} = (7 - 2x + 3x^2) \) N on a 2 kg mass as it displaces from \( x = 0 \) m to \( x = 5 \) m, we will follow these steps:

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

### Step 2: Set up the integral
In this case, the force is given by:
\[ F(x) = 7 - 2x + 3x^2 \]
We need to calculate the work done from \( x = 0 \) to \( x = 5 \):
\[ W = \int_{0}^{5} (7 - 2x + 3x^2) \, dx \]

### Step 3: Calculate the integral
Now we will compute the integral:
\[ W = \int_{0}^{5} (7 - 2x + 3x^2) \, dx \]
We can integrate term by term:
1. The integral of \( 7 \) is \( 7x \).
2. The integral of \( -2x \) is \( -x^2 \).
3. The integral of \( 3x^2 \) is \( x^3 \).

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

### Step 4: Evaluate the definite integral
Now we will evaluate this expression at the limits \( x = 5 \) and \( x = 0 \):
1. At \( x = 5 \):
\[ W(5) = 7(5) - (5)^2 + (5)^3 = 35 - 25 + 125 = 135 \]
2. At \( x = 0 \):
\[ W(0) = 7(0) - (0)^2 + (0)^3 = 0 \]

Now, substituting these values into the expression for work done:
\[ W = W(5) - W(0) = 135 - 0 = 135 \, \text{J} \]

### Final Answer
The work done by the force on the mass as it moves from \( x = 0 \) m to \( x = 5 \) m is:
\[ \boxed{135 \, \text{J}} \]