A body of mass 2kg collides with a horizontal weight spring of force constant `4Nm^(-1)`. The body compresses the spring by 1m from rest position. Find the speed of the block at the instant of collision ? Given that the coefficient of kinetic friction between the body and the surface is `0.1`.

Explanation

To find the speed of the block at the instant of collision with the spring, we can use the principle of conservation of energy, taking into account the work done against friction.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the body, \( m = 2 \, \text{kg} \)
- Spring constant, \( k = 4 \, \text{N/m} \)
- Compression of the spring, \( x = 1 \, \text{m} \)
- Coefficient of kinetic friction, \( \mu = 0.1 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Initial Kinetic Energy (KE):**
The initial kinetic energy of the block before it collides with the spring is given by:
\[ KE = \frac{1}{2} m v^2 \]
where \( v \) is the speed of the block at the instant of collision.

3. **Calculate the Potential Energy (PE) Stored in the Spring:**
The potential energy stored in the spring when it is compressed by \( x \) is given by:
\[ PE = \frac{1}{2} k x^2 \]
Substituting the values:
\[ PE = \frac{1}{2} \times 4 \times (1)^2 = 2 \, \text{J} \]

4. **Calculate the Work Done Against Friction (W_f):**
The work done against friction can be calculated using:
\[ W_f = \text{friction force} \times \text{displacement} \]
The friction force is given by:
\[ F_f = \mu m g = 0.1 \times 2 \times 10 = 2 \, \text{N} \]
Therefore, the work done against friction when the block compresses the spring by \( x \):
\[ W_f = F_f \times x = 2 \times 1 = 2 \, \text{J} \]

5. **Apply the Energy Conservation Principle:**
The initial kinetic energy is converted into the potential energy of the spring and the work done against friction:
\[ KE = PE + W_f \]
Substituting the known values:
\[ \frac{1}{2} m v^2 = 2 + 2 \]
\[ \frac{1}{2} \times 2 v^2 = 4 \]
Simplifying gives:
\[ v^2 = 4 \]
\[ v = 2 \, \text{m/s} \]

### Final Answer:
The speed of the block at the instant of collision is \( 2 \, \text{m/s} \).

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