A block of ice at `0^(@)C` whose mass is initially 50.0 kg slides along a horizontal surface, starting at a speed of 5.38 m/s and finally coming tor est after travelling 28.3 meters. The mass of ice melted as a result of the friction between the block and the surface will be

Options

A. 2.16 g
B. 4.0 g
C. 1 g
D. 50 g

Answer

To solve the problem of how much ice melts due to the friction between the block and the surface, we can follow these steps:

### Step 1: Calculate the initial kinetic energy of the ice block.
The formula for kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} mv^2 \]
Where:
- \( m = 50.0 \, \text{kg} \) (mass of the ice block)
- \( v = 5.38 \, \text{m/s} \) (initial speed)

Substituting the values:
\[ KE = \frac{1}{2} \times 50.0 \, \text{kg} \times (5.38 \, \text{m/s})^2 \]
Calculating \( (5.38)^2 \):
\[ (5.38)^2 = 28.9444 \]
Now substituting back:
\[ KE = \frac{1}{2} \times 50.0 \times 28.9444 = 25.0 \times 28.9444 = 723.61 \, \text{J} \]

### Step 2: Relate the kinetic energy to the heat generated due to friction.
The kinetic energy lost by the block is converted into heat energy (Q), which is used to melt the ice. Therefore:
\[ Q = KE = 723.61 \, \text{J} \]

### Step 3: Use the latent heat of fusion to find the mass of ice melted.
The heat required to melt ice can be expressed as:
\[ Q = m \cdot L \]
Where:
- \( m \) is the mass of ice melted (in kg)
- \( L = 334 \, \text{J/kg} \) (latent heat of fusion for ice)

Rearranging the equation to find \( m \):
\[ m = \frac{Q}{L} = \frac{723.61 \, \text{J}}{334 \, \text{J/kg}} \]
Calculating \( m \):
\[ m = 2.16 \, \text{kg} \]

### Step 4: Convert the mass from kg to grams.
Since the question asks for the mass in grams:
\[ m = 2.16 \, \text{kg} \times 1000 \, \text{g/kg} = 2160 \, \text{g} \]

### Final Answer:
The mass of ice melted as a result of the friction between the block and the surface is **2160 grams**.

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