A particle moves along X-axis from `x=0` to `x=1` m under the influence of a force given by `F=3x^(2)+2x-10.` Work done in the process is

Options

A. `+4J`
B. `-4J`
C. `+8J`
D. `-8J`

Answer

To find the work done by the force \( F = 3x^2 + 2x - 10 \) as a particle moves from \( x = 0 \) m to \( x = 1 \) m, we will use the formula for work done by a variable force:

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

### Step 1: Set up the integral
We need to evaluate the integral of the force from \( x = 0 \) to \( x = 1 \):

\[ W = \int_{0}^{1} (3x^2 + 2x - 10) \, dx \]

### Step 2: Integrate the function
Now, we can integrate the function term by term:

\[ W = \int (3x^2) \, dx + \int (2x) \, dx + \int (-10) \, dx \]

Calculating each integral separately:

1. \(\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3\)
2. \(\int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2\)
3. \(\int -10 \, dx = -10x\)

Combining these results, we have:

\[ W = \left[ x^3 + x^2 - 10x \right]_{0}^{1} \]

### Step 3: Evaluate the definite integral
Now, we will evaluate the expression at the limits \( x = 1 \) and \( x = 0 \):

1. At \( x = 1 \):
\[ W(1) = 1^3 + 1^2 - 10 \cdot 1 = 1 + 1 - 10 = -8 \]

2. At \( x = 0 \):
\[ W(0) = 0^3 + 0^2 - 10 \cdot 0 = 0 \]

### Step 4: Calculate the work done
Now, we subtract the value at \( x = 0 \) from the value at \( x = 1 \):

\[ W = W(1) - W(0) = -8 - 0 = -8 \, \text{Joules} \]

Thus, the work done by the force as the particle moves from \( x = 0 \) m to \( x = 1 \) m is:

\[ \boxed{-8 \, \text{Joules}} \]