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

Explanation

To find the final kinetic energy of the block, we will follow these steps:

### Step 1: Identify the given values
- Mass of the block, \( m = 10 \, \text{kg} \)
- Initial speed, \( v_i = 10 \, \text{m/s} \)
- Initial position, \( x_1 = 20 \, \text{m} \)
- Final position, \( x_2 = 30 \, \text{m} \)
- Retarding force, \( F(x) = -0.1x \, \text{J/m} \)

### Step 2: Calculate the initial kinetic energy (KE_initial)
The initial kinetic energy can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m v_i^2 \]
Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 10 \, \text{kg} \times (10 \, \text{m/s})^2 = \frac{1}{2} \times 10 \times 100 = 500 \, \text{J} \]

### Step 3: Calculate the work done by the retarding force
The work done by the force can be calculated using the integral of the force over the distance from \( x_1 \) to \( x_2 \):
\[ W = \int_{x_1}^{x_2} F(x) \, dx = \int_{20}^{30} -0.1x \, dx \]
Calculating the integral:
\[ W = -0.1 \int_{20}^{30} x \, dx = -0.1 \left[ \frac{x^2}{2} \right]_{20}^{30} = -0.1 \left( \frac{30^2}{2} - \frac{20^2}{2} \right) \]
Calculating the values:
\[ = -0.1 \left( \frac{900}{2} - \frac{400}{2} \right) = -0.1 \left( 450 - 200 \right) = -0.1 \times 250 = -25 \, \text{J} \]

### Step 4: Calculate the final kinetic energy (KE_final)
Using the work-energy principle, the final kinetic energy can be calculated as:
\[ KE_{\text{final}} = KE_{\text{initial}} + W \]
Substituting the values:
\[ KE_{\text{final}} = 500 \, \text{J} - 25 \, \text{J} = 475 \, \text{J} \]

### Final Answer
The final kinetic energy of the block is \( 475 \, \text{J} \).

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