A block of ice of mass 50kg is sliding on a horizontal plane. It starts with speed `5ms^(-1)` and stops after moving through some distance. The mass of ice that has melted due to friction between the block and the surface is (assuming that no energy is lost and the surface is (assuming that no energy is lost and latent heat of fusion of ice is 80 `cal g^(-1)`)

Options

A. 2.86 g
B. 3.86 g
C. 0.86 g
D. 1.86 g

Explanation

To solve the problem, we need to determine the mass of ice that melts due to the friction between the sliding block of ice and the horizontal surface. We will use the principles of energy conservation, where the kinetic energy lost by the ice block is equal to the heat gained by the ice to melt.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the ice block, \( m = 50 \, \text{kg} \)
- Initial speed of the ice block, \( v = 5 \, \text{m/s} \)
- Latent heat of fusion of ice, \( L = 80 \, \text{cal/g} \)

2. **Convert the latent heat of fusion from calories to joules:**
- We know that \( 1 \, \text{cal} = 4.2 \, \text{J} \).
- Therefore, \( L = 80 \, \text{cal/g} \times 4.2 \, \text{J/cal} = 336 \, \text{J/g} \).

3. **Calculate the initial kinetic energy (KE) of the ice block:**
- The formula for kinetic energy is given by:
\[ KE = \frac{1}{2} m v^2 \]
- Substituting the values:
\[ KE = \frac{1}{2} \times 50 \, \text{kg} \times (5 \, \text{m/s})^2 = \frac{1}{2} \times 50 \times 25 = 625 \, \text{J} \]

4. **Set up the equation for heat gained by the ice:**
- Let \( m' \) be the mass of ice melted (in grams).
- The heat gained by the ice when it melts is given by:
\[ Q = m' \times L \]
- Substituting the value of \( L \):
\[ Q = m' \times 336 \, \text{J/g} \]

5. **Equate the heat lost to the heat gained:**
- According to the principle of calorimetry:
\[ KE = Q \]
- Therefore:
\[ 625 \, \text{J} = m' \times 336 \, \text{J/g} \]

6. **Solve for \( m' \):**
- Rearranging the equation gives:
\[ m' = \frac{625 \, \text{J}}{336 \, \text{J/g}} \approx 1.86 \, \text{g} \]

7. **Conclusion:**
- The mass of ice that has melted due to friction is approximately \( 1.86 \, \text{g} \).