A particle of mass `20gm` is moving with velocity `1m//s`. It penetrates `20 cm` wooden block (fixed) with average force `2.5xx10^(-2)N`. Find out speed particle when it come out from blocl.

Options

A. `(1)/(sqrt(3))m//s`
B. `(1)/(sqrt(5))m//s`
C. `(1)/(sqrt(2))m//s`
D. `(1)/(sqrt(7))m//s`

Answer

To solve the problem, we will apply 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 Given Data:**
- Mass of the particle, \( m = 20 \, \text{g} = 20 \times 10^{-3} \, \text{kg} = 0.02 \, \text{kg} \)
- Initial velocity of the particle, \( v_i = 1 \, \text{m/s} \)
- Distance penetrated into the block, \( x = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Average force exerted by the block, \( F = 2.5 \times 10^{-2} \, \text{N} \)

2. **Apply the Work-Energy Theorem:**
The work done by the force on the particle is equal to the change in kinetic energy:
\[ W = -F \cdot x = \Delta KE \]
where \( \Delta KE = KE_f - KE_i \) and \( KE = \frac{1}{2} m v^2 \).

3. **Calculate Initial Kinetic Energy:**
\[ KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 0.02 \times (1)^2 = 0.01 \, \text{J} \]

4. **Calculate Work Done:**
\[ W = -F \cdot x = - (2.5 \times 10^{-2}) \cdot (0.2) = -0.005 \, \text{J} \]

5. **Set Up the Equation:**
\[ -0.005 = KE_f - 0.01 \]
Rearranging gives:
\[ KE_f = 0.01 - 0.005 = 0.005 \, \text{J} \]

6. **Calculate Final Kinetic Energy:**
\[ KE_f = \frac{1}{2} m v_f^2 \]
Setting this equal to the calculated final kinetic energy:
\[ 0.005 = \frac{1}{2} \times 0.02 \times v_f^2 \]

7. **Solve for Final Velocity \( v_f \):**
\[ 0.005 = 0.01 v_f^2 \]
\[ v_f^2 = \frac{0.005}{0.01} = 0.5 \]
\[ v_f = \sqrt{0.5} = \frac{1}{\sqrt{2}} \, \text{m/s} \]

### Final Answer:
The speed of the particle when it comes out from the block is \( \frac{1}{\sqrt{2}} \, \text{m/s} \).

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