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 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: Identify the Initial Kinetic Energy
The initial kinetic energy (KE_initial) of the block can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m v^2 \]
where:
- \( m = 20 \, \text{kg} \) (mass of the block)
- \( v = 10 \, \text{m/s} \) (initial speed)

Calculating:
\[ KE_{\text{initial}} = \frac{1}{2} \times 20 \times (10)^2 = \frac{1}{2} \times 20 \times 100 = 1000 \, \text{J} \]

### Step 2: Calculate 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 integral of the force over the distance:
\[ W = \int_{20}^{30} F \, dx \]
Given that the force \( F = -0.1x \):
\[ W = \int_{20}^{30} -0.1x \, dx \]

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

### Step 3: Calculate the Final Kinetic Energy
The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy:
\[ W = KE_{\text{final}} - KE_{\text{initial}} \]
Rearranging gives:
\[ KE_{\text{final}} = KE_{\text{initial}} + W \]
Substituting the values we calculated:
\[ KE_{\text{final}} = 1000 \, \text{J} - 25 \, \text{J} = 975 \, \text{J} \]

### Final Answer
The final kinetic energy of the block is:
\[ \boxed{975 \, \text{J}} \]

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