A variable force F acts along the x - axis given by `F=(3x^2)-2x+1N`. The work done by the force when a particle of mass 100 g moves from x = 50 cm to x = 100 cm is

Options

A. (a)0.625 J
B. (b)6.25 J
C. (c)0.0625 J
D. (d)62.5 J

Answer

To solve the problem of finding the work done by the variable force \( F = 3x^2 - 2x + 1 \) when a particle of mass 100 g moves from \( x = 50 \) cm to \( x = 100 \) cm, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Units**:
- The mass of the particle is given as 100 g, which is \( 0.1 \) kg (since \( 1 \) g = \( 0.001 \) kg).
- The distance needs to be converted from centimeters to meters:
- \( x = 50 \) cm = \( 0.5 \) m
- \( x = 100 \) cm = \( 1 \) m

2. **Set Up the Work Done Integral**:
- The work done \( W \) by a variable force is given by the integral of the force over the displacement:
\[ W = \int_{x_1}^{x_2} F \, dx \]
- Here, \( F = 3x^2 - 2x + 1 \), \( x_1 = 0.5 \), and \( x_2 = 1 \).

3. **Integrate the Force Function**:
- We need to calculate:
\[ W = \int_{0.5}^{1} (3x^2 - 2x + 1) \, dx \]
- The integral can be calculated as follows:
\[ W = \left[ x^3 - x^2 + x \right]_{0.5}^{1} \]

4. **Evaluate the Integral at the Limits**:
- First, evaluate at the upper limit \( x = 1 \):
\[ W(1) = 1^3 - 1^2 + 1 = 1 - 1 + 1 = 1 \]
- Now evaluate at the lower limit \( x = 0.5 \):
\[ W(0.5) = (0.5)^3 - (0.5)^2 + (0.5) = \frac{1}{8} - \frac{1}{4} + \frac{1}{2} \]
- Converting to a common denominator (8):
\[ W(0.5) = \frac{1}{8} - \frac{2}{8} + \frac{4}{8} = \frac{1 - 2 + 4}{8} = \frac{3}{8} \]

5. **Calculate the Work Done**:
- Now, subtract the two results:
\[ W = W(1) - W(0.5) = 1 - \frac{3}{8} = \frac{8}{8} - \frac{3}{8} = \frac{5}{8} \]
- Therefore, the work done is:
\[ W = 0.625 \, \text{J} \]

### Final Answer:
The work done by the force when the particle moves from \( x = 50 \) cm to \( x = 100 \) cm is \( 0.625 \, \text{J} \).

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