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 need to find the increase in temperature of a copper cube sliding down a rough inclined plane. Here are the steps to arrive at the solution:

### Step 1: Understand the Given Data
- Mass of the copper cube (m) = 200 g = 0.2 kg (conversion from grams to kilograms)
- Inclination angle (θ) = 37 degrees
- Distance slid down the incline (L) = 60 cm = 0.6 m (conversion from centimeters to meters)
- Specific heat capacity of copper (S) = 420 J/(kg·K)

### Step 2: Calculate the Change in Potential Energy
The change in potential energy (ΔPE) as the block slides down can be calculated using the formula:
\[ \Delta PE = m \cdot g \cdot h \]
where \( h \) is the vertical height descended, which can be calculated as:
\[ h = L \cdot \sin(θ) \]
Substituting the values:
- \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \)
- \( h = 0.6 \cdot \sin(37^\circ) \)

Using \( \sin(37^\circ) \approx 0.6 \):
\[ h = 0.6 \cdot 0.6 = 0.36 \, \text{m} \]

Now, substituting the values into the potential energy formula:
\[ \Delta PE = 0.2 \cdot 10 \cdot 0.36 = 0.72 \, \text{J} \]

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

### Step 4: Use the Formula for Thermal Energy
The thermal energy gained is given by:
\[ \Delta Q = m \cdot S \cdot \Delta T \]
where \( \Delta T \) is the increase in temperature.

Setting the two expressions for energy equal:
\[ 0.72 = 0.2 \cdot 420 \cdot \Delta T \]

### Step 5: Solve for Increase in Temperature (ΔT)
Rearranging the equation to find \( \Delta T \):
\[ \Delta T = \frac{0.72}{0.2 \cdot 420} \]
Calculating the denominator:
\[ 0.2 \cdot 420 = 84 \]
Now substituting back:
\[ \Delta T = \frac{0.72}{84} \approx 0.00857 \, \text{K} \]

### Step 6: Final Answer
Thus, the increase in temperature of the copper block is approximately:
\[ \Delta T \approx 0.00857 \, \text{K} \]