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

Options

A. 975 joule
B. 450 joule
C. 275 joule
D. 950 joule

Explanation

To find the final kinetic energy of the block, we can 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 the Initial Conditions**:
- Mass of the block, \( m = 20 \, \text{kg} \)
- Initial speed, \( v_i = 10 \, \text{m/s} \)
- Initial kinetic energy, \( KE_i = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 20 \times (10)^2 = 1000 \, \text{J} \)

2. **Determine the Retarding Force**:
- The retarding force is given by \( F = -0.1x \, \text{N} \).

3. **Calculate the Work Done by the Force**:
- The work done \( W \) by the force as the block moves from \( x = 20 \, \text{m} \) to \( x = 30 \, \text{m} \) can be calculated using the integral:
\[ W = \int_{20}^{30} F \, dx = \int_{20}^{30} (-0.1x) \, dx \]

4. **Evaluate the Integral**:
- Calculate the integral:
\[ W = -0.1 \int_{20}^{30} x \, dx = -0.1 \left[ \frac{x^2}{2} \right]_{20}^{30} \]
- Evaluating the limits:
\[ W = -0.1 \left( \frac{30^2}{2} - \frac{20^2}{2} \right) = -0.1 \left( \frac{900}{2} - \frac{400}{2} \right) = -0.1 \left( 450 - 200 \right) = -0.1 \times 250 = -25 \, \text{J} \]

5. **Apply the Work-Energy Theorem**:
- According to the work-energy theorem:
\[ W = KE_f - KE_i \]
- Rearranging gives:
\[ KE_f = KE_i + W = 1000 \, \text{J} - 25 \, \text{J} = 975 \, \text{J} \]

6. **Conclusion**:
- The final kinetic energy of the block is \( KE_f = 975 \, \text{J} \).

### Final Answer:
The final kinetic energy of the block is **975 Joules**.