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 solve the problem, we will follow these steps:

### Step 1: Identify the initial kinetic energy (KE_initial)
The initial kinetic energy (KE_initial) of the block can be calculated using the formula:
\[ KE_{initial} = \frac{1}{2} m v^2 \]
where:
- \( m = 10 \, \text{kg} \) (mass of the block)
- \( v = 10 \, \text{m/s} \) (initial speed)

Substituting the values:
\[ KE_{initial} = \frac{1}{2} \times 10 \times (10)^2 = \frac{1}{2} \times 10 \times 100 = 500 \, \text{J} \]

### Step 2: Calculate the work done by the retarding force
The retarding force is given by:
\[ F = 0.1x \, \text{J/m} \]
To find the work done (W) by this force as the block moves from \( x = 20 \, \text{m} \) to \( x = 30 \, \text{m} \), we need to integrate the force over the distance:
\[ W = \int_{x_1}^{x_2} F \, dx = \int_{20}^{30} 0.1x \, dx \]

Calculating the integral:
\[ W = 0.1 \int_{20}^{30} x \, dx = 0.1 \left[ \frac{x^2}{2} \right]_{20}^{30} \]
\[ = 0.1 \left( \frac{30^2}{2} - \frac{20^2}{2} \right) = 0.1 \left( \frac{900}{2} - \frac{400}{2} \right) \]
\[ = 0.1 \left( 450 - 200 \right) = 0.1 \times 250 = 25 \, \text{J} \]

### Step 3: Determine the final kinetic energy (KE_final)
Using the work-energy principle, the final kinetic energy can be calculated as:
\[ KE_{final} = KE_{initial} + W \]
Substituting the values we found:
\[ KE_{final} = 500 \, \text{J} - 25 \, \text{J} = 475 \, \text{J} \]

### Final Answer
The final kinetic energy of the block is:
\[ \boxed{475 \, \text{J}} \]