A block of mass `100g` slides on a rough horizontal surface .If the speed of the block decreases from `10ms^(-1) to 5ms^(-1)`, find the thermal energy developed in the process

Explanation

To find the thermal energy developed when a block slides on a rough horizontal surface and its speed decreases, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the block, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) (convert grams to kilograms by dividing by 1000).
- Initial speed, \( v_i = 10 \, \text{m/s} \).
- Final speed, \( v_f = 5 \, \text{m/s} \).

2. **Calculate the Initial and Final Kinetic Energy:**
- The formula for kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} m v^2 \]
- Calculate the initial kinetic energy (\( KE_i \)):
\[ KE_i = \frac{1}{2} \times 0.1 \, \text{kg} \times (10 \, \text{m/s})^2 = \frac{1}{2} \times 0.1 \times 100 = 5 \, \text{J} \]
- Calculate the final kinetic energy (\( KE_f \)):
\[ KE_f = \frac{1}{2} \times 0.1 \, \text{kg} \times (5 \, \text{m/s})^2 = \frac{1}{2} \times 0.1 \times 25 = 1.25 \, \text{J} \]

3. **Calculate the Loss of Kinetic Energy:**
- The loss of kinetic energy is the difference between the initial and final kinetic energy:
\[ \Delta KE = KE_i - KE_f = 5 \, \text{J} - 1.25 \, \text{J} = 3.75 \, \text{J} \]

4. **Conclusion:**
- The thermal energy developed due to the loss of kinetic energy is:
\[ TE = \Delta KE = 3.75 \, \text{J} \]

### Final Answer:
The thermal energy developed in the process is **3.75 Joules**.

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