A block of `10 kg` mass is thrown on a rough surface having friction coefficient `0.3` with an initial speed of `5` m/s. Find the amount by which the internal energy of the blocked and the surface will increase. If this block is seen from a frame moving at a speed of `5` m/s in the direction of its velocity it is observed that the block is gently put on the rough surface moving in opposite direction at speed `5` m/s. Find the gain kinetic energy of the block as seen from this frame.

Answer

To solve the problem step by step, we will break it into two parts as described in the question.

### Part 1: Change in Internal Energy

1. **Identify the initial kinetic energy of the block:**
The formula for kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass and \( v \) is the velocity.

Given:
- Mass \( m = 10 \, \text{kg} \)
- Initial velocity \( v = 5 \, \text{m/s} \)

Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 10 \times (5)^2 = \frac{1}{2} \times 10 \times 25 = 125 \, \text{J} \]

2. **Determine the final kinetic energy of the block:**
When the block comes to rest, its final velocity is \( 0 \, \text{m/s} \). Therefore, the final kinetic energy is:
\[ KE_{\text{final}} = \frac{1}{2} \times 10 \times (0)^2 = 0 \, \text{J} \]

3. **Calculate the change in kinetic energy:**
The change in kinetic energy (\( \Delta KE \)) is given by:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 0 - 125 = -125 \, \text{J} \]

4. **Relate the change in kinetic energy to the change in internal energy:**
The loss of kinetic energy is converted into heat energy due to friction, which increases the internal energy of the block and the surface. Thus:
\[ \Delta E = -\Delta KE = 125 \, \text{J} \]

### Part 2: Gain in Kinetic Energy from Another Frame

1. **Analyze the situation from a moving frame:**
In this frame, both the observer and the block are moving at the same speed of \( 5 \, \text{m/s} \) in the same direction. Therefore, the relative velocity of the block with respect to the observer is:
\[ v_{\text{relative}} = v_{\text{block}} - v_{\text{frame}} = 5 - 5 = 0 \, \text{m/s} \]

2. **Determine the initial kinetic energy in the moving frame:**
Since the block appears at rest in this frame, its initial kinetic energy is:
\[ KE_{\text{initial, frame}} = 0 \, \text{J} \]

3. **Calculate the final kinetic energy in the moving frame:**
If the surface is moving in the opposite direction at \( 5 \, \text{m/s} \), the block will eventually have a velocity of \( -5 \, \text{m/s} \) relative to the observer. Therefore, the final kinetic energy is:
\[ KE_{\text{final, frame}} = \frac{1}{2} \times 10 \times (-5)^2 = \frac{1}{2} \times 10 \times 25 = 125 \, \text{J} \]

4. **Calculate the gain in kinetic energy:**
The gain in kinetic energy (\( \Delta KE_{\text{frame}} \)) is:
\[ \Delta KE_{\text{frame}} = KE_{\text{final, frame}} - KE_{\text{initial, frame}} = 125 - 0 = 125 \, \text{J} \]

### Final Answers
- The increase in internal energy of the block and surface is **125 J**.
- The gain in kinetic energy of the block as seen from the moving frame is **125 J**.