A bolck of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retardeng force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy 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 can follow these steps:

### Step 1: Calculate the Initial Kinetic Energy
The initial kinetic energy (KE_initial) of the block can be calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( m = 10 \, \text{kg} \) and \( v = 10 \, \text{m/s} \).

\[ KE_{\text{initial}} = \frac{1}{2} \times 10 \, \text{kg} \times (10 \, \text{m/s})^2 \]
\[ KE_{\text{initial}} = \frac{1}{2} \times 10 \times 100 = 500 \, \text{J} \]

### Step 2: Determine the Work Done by the Retarding Force
The work done (W) by the retarding force as the block moves from \( x = 20 \, \text{m} \) to \( x = 30 \, \text{m} \) can be calculated using the formula:
\[ W = \int_{x_1}^{x_2} F \, dx \]
Given that the force \( F = -0.1x \, \text{J/m} \), we can set up the integral:
\[ W = \int_{20}^{30} -0.1x \, dx \]

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: Calculate the Final Kinetic Energy
Using the work-energy theorem, the change in kinetic energy is equal to the work done:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = W \]
Substituting the values:
\[ KE_{\text{final}} - 500 \, \text{J} = -25 \, \text{J} \]
\[ KE_{\text{final}} = 500 \, \text{J} - 25 \, \text{J} \]
\[ KE_{\text{final}} = 475 \, \text{J} \]

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

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