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 it slides down through `60cm`. Specific heat capacity of copper `= 420J kg^(-1) K^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the Given Values
- Mass of the copper cube, \( m = 200 \, \text{g} = 0.2 \, \text{kg} \)
- Length slid down the incline, \( L = 60 \, \text{cm} = 0.6 \, \text{m} \)
- Angle of inclination, \( \theta = 37^\circ \)
- Specific heat capacity of copper, \( s = 420 \, \text{J kg}^{-1} \text{K}^{-1} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Change in Potential Energy
The change in potential energy as the block slides down can be calculated using the formula:
\[ \Delta PE = mgh \]
where \( h \) is the vertical height descended. The height \( h \) can be found using the sine of the angle:
\[ h = L \sin(\theta) \]
Substituting the values:
\[ h = 0.6 \sin(37^\circ) = 0.6 \times \frac{3}{5} = 0.36 \, \text{m} \]
Now substituting \( h \) back into the potential energy formula:
\[ \Delta PE = 0.2 \times 10 \times 0.36 = 0.72 \, \text{J} \]

### Step 3: Relate Change in Potential Energy to Thermal Energy
According to the problem, the change in potential energy is converted into thermal energy:
\[ \Delta PE = \Delta Q \]
where \( \Delta Q \) is the thermal energy gained by the block.

### Step 4: Use the Formula for Thermal Energy
The thermal energy gained by the block can also be expressed as:
\[ \Delta Q = ms\Delta T \]
where \( \Delta T \) is the change in temperature. Setting the two expressions for thermal energy equal gives:
\[ ms\Delta T = 0.72 \]

### Step 5: Solve for Change in Temperature
Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{0.72}{ms} \]
Substituting the known values:
\[ \Delta T = \frac{0.72}{0.2 \times 420} \]
Calculating:
\[ \Delta T = \frac{0.72}{84} = 0.00857 \, \text{K} \approx 8.57 \, \text{mK} \]

### Final Answer
The increase in the temperature of the copper block as it slides down through 60 cm is approximately \( 8.57 \, \text{mK} \).

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