A copper cube of mass `200g` slides down an a rough inclined plane of inclination `37^(@)` at a constant speed Assume that any loss in mechanical energy goes into the copper block as thermal energy .Find the increase in the temperature of the block as if slides down through `60cm` Specific head capacity of copper `= 420J kg^(-1) K^(-1)`

Explanation

To solve the problem step by step, we will follow the principles of energy conservation and the definition of specific heat capacity.

### Step 1: Identify the given data
- Mass of the copper cube, \( m = 200 \, \text{g} = 0.2 \, \text{kg} \)
- Distance slid down the incline, \( L = 60 \, \text{cm} = 0.6 \, \text{m} \)
- Angle of inclination, \( \theta = 37^\circ \)
- Specific heat capacity of copper, \( c = 420 \, \text{J/kg/K} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate the change in potential energy
The change in potential energy as the cube slides down the incline can be calculated using the formula:
\[ \Delta PE = mgh \]
where \( h \) is the height change. The height \( h \) can be calculated using:
\[ h = L \sin(\theta) \]
Substituting the values:
\[ h = 0.6 \, \text{m} \cdot \sin(37^\circ) \]
Using \( \sin(37^\circ) \approx 0.6 \):
\[ h \approx 0.6 \cdot 0.6 = 0.36 \, \text{m} \]
Now, substituting \( h \) back into the potential energy formula:
\[ \Delta PE = 0.2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot 0.36 \, \text{m} \]
Calculating this gives:
\[ \Delta PE \approx 0.2 \cdot 9.8 \cdot 0.36 \approx 0.7056 \, \text{J} \]

### Step 3: Relate the change in potential energy to thermal energy
Since the cube slides down at a constant speed, the loss in potential energy is converted into thermal energy gained by the copper block:
\[ \Delta PE = \Delta Q \]
where \( \Delta Q = mc\Delta T \). Therefore,
\[ 0.7056 \, \text{J} = 0.2 \, \text{kg} \cdot 420 \, \text{J/kg/K} \cdot \Delta T \]

### Step 4: Solve for the increase in temperature \( \Delta T \)
Rearranging the equation to find \( \Delta T \):
\[ \Delta T = \frac{0.7056 \, \text{J}}{0.2 \cdot 420} \]
Calculating the denominator:
\[ 0.2 \cdot 420 = 84 \]
Now substituting back:
\[ \Delta T = \frac{0.7056}{84} \approx 0.0084 \, \text{K} \]

### Step 5: Final result
The increase in temperature of the copper block is:
\[ \Delta T \approx 0.0084 \, \text{K} \text{ or } 8.4 \times 10^{-3} \, \text{K} \]