A block of mass 10 kg, moving in x-direction with a constant speed of `10ms^(-1)`, is subjected to a retarding force `F=0.1xxJ//m` during its travel from x=20 m to 30 m. Its final KE will be

Options

A. 475 J
B. 450 J
C. 275 J
D. 250 J

Explanation

To find the final kinetic energy of the block, we will use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the block, \( m = 10 \, \text{kg} \)
- Initial speed, \( v = 10 \, \text{m/s} \)
- Initial position, \( x_1 = 20 \, \text{m} \)
- Final position, \( x_2 = 30 \, \text{m} \)
- Retarding force, \( F = -0.1x \, \text{J/m} \) (negative because it is a retarding force)

2. **Calculate Initial Kinetic Energy (KE):**
\[ KE_{\text{initial}} = \frac{1}{2}mv^2 = \frac{1}{2} \times 10 \, \text{kg} \times (10 \, \text{m/s})^2 = 500 \, \text{J} \]

3. **Calculate Work Done by the Retarding Force:**
- The work done \( W \) by the force as the block moves from \( x = 20 \, \text{m} \) to \( x = 30 \, \text{m} \) is given by:
\[ W = \int_{x_1}^{x_2} F \, dx = \int_{20}^{30} (-0.1x) \, dx \]
- Evaluating the integral:
\[ W = -0.1 \int_{20}^{30} x \, dx = -0.1 \left[ \frac{x^2}{2} \right]_{20}^{30} \]
\[ = -0.1 \left( \frac{30^2}{2} - \frac{20^2}{2} \right) = -0.1 \left( \frac{900}{2} - \frac{400}{2} \right) \]
\[ = -0.1 \left( 450 - 200 \right) = -0.1 \times 250 = -25 \, \text{J} \]

4. **Calculate Final Kinetic Energy:**
- According to the work-energy theorem:
\[ W = KE_{\text{final}} - KE_{\text{initial}} \]
Rearranging gives:
\[ KE_{\text{final}} = W + KE_{\text{initial}} = -25 \, \text{J} + 500 \, \text{J} = 475 \, \text{J} \]

5. **Final Result:**
The final kinetic energy of the block is:
\[ KE_{\text{final}} = 475 \, \text{J} \]