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

### Step 1: Set up the work integral
The work done \( W \) by a force \( F \) when moving along a path is given by the integral:
\[ W = \int_{x_1}^{x_2} F \, dx \]
Here, \( x_1 = 0 \) m and \( x_2 = 1 \) m.

### Step 2: Substitute the force into the integral
Substituting the expression for \( F \):
\[ W = \int_{0}^{1} (3x^2 + 2x - 10) \, dx \]

### Step 3: Integrate the expression
Now, we will integrate each term separately:
\[ W = \int_{0}^{1} 3x^2 \, dx + \int_{0}^{1} 2x \, dx - \int_{0}^{1} 10 \, dx \]

Calculating each integral:
1. \(\int 3x^2 \, dx = x^3\)
2. \(\int 2x \, dx = x^2\)
3. \(\int 10 \, dx = 10x\)

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

### Step 4: Evaluate the definite integrals
Now we evaluate the limits:
1. For \( x^3 \) from 0 to 1:
\[ \left[ x^3 \right]_{0}^{1} = 1^3 - 0^3 = 1 - 0 = 1 \]
2. For \( x^2 \) from 0 to 1:
\[ \left[ x^2 \right]_{0}^{1} = 1^2 - 0^2 = 1 - 0 = 1 \]
3. For \( 10x \) from 0 to 1:
\[ \left[ 10x \right]_{0}^{1} = 10(1) - 10(0) = 10 - 0 = 10 \]

### Step 5: Combine the results
Now, substituting these results back into the expression for work:
\[ W = 1 + 1 - 10 = 2 - 10 = -8 \, \text{J} \]

### Final Answer
The work done in the process is:
\[ \boxed{-8 \, \text{J}} \]