A positon dependent force ,`F=8-4x+3x^(2)N` acts on a small body of mass 2 kg and displaces it from x=0 to x=5 m. The work done in joule is

Options

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

Explanation

To find the work done by the position-dependent force \( F = 8 - 4x + 3x^2 \) on a small body of mass 2 kg as it displaces from \( x = 0 \) to \( x = 5 \) m, we will follow these steps:

### Step 1: Understand the formula for work done
The work done \( W \) by a force \( F \) when moving an object through a displacement \( dx \) is given by the integral of the force with respect to displacement:
\[ W = \int F \, dx \]

### Step 2: Set up the integral
Given the force \( F = 8 - 4x + 3x^2 \), we need to integrate this force from \( x = 0 \) to \( x = 5 \):
\[ W = \int_{0}^{5} (8 - 4x + 3x^2) \, dx \]

### Step 3: Perform the integration
We will integrate each term separately:
1. The integral of \( 8 \) with respect to \( x \) is \( 8x \).
2. The integral of \( -4x \) is \( -2x^2 \).
3. The integral of \( 3x^2 \) is \( x^3 \).

Thus, the integral becomes:
\[ W = \left[ 8x - 2x^2 + x^3 \right]_{0}^{5} \]

### Step 4: Evaluate the integral at the limits
Now we will evaluate the expression at the upper limit \( x = 5 \) and the lower limit \( x = 0 \):
1. At \( x = 5 \):
\[ W(5) = 8(5) - 2(5^2) + (5^3) = 40 - 50 + 125 = 115 \]
2. At \( x = 0 \):
\[ W(0) = 8(0) - 2(0^2) + (0^3) = 0 \]

### Step 5: Calculate the work done
Now subtract the value at the lower limit from the value at the upper limit:
\[ W = W(5) - W(0) = 115 - 0 = 115 \text{ joules} \]

### Final Answer
The work done by the force is \( \boxed{115} \) joules.