A block of mass 250 g slides down an incline of inclination `37^0` with a uniform speed. Find the work done against the friction as the block slides through 1.0 m.

Explanation

To solve the problem of finding the work done against friction as a block slides down an incline, we will follow these steps:

### Step 1: Identify the given data
- Mass of the block (m) = 250 g = 0.25 kg (since 1 g = 0.001 kg)
- Inclination angle (θ) = 37°
- Displacement (d) = 1.0 m

### Step 2: Calculate the gravitational force acting on the block
The weight (W) of the block can be calculated using the formula:
\[ W = m \cdot g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).

Calculating the weight:
\[ W = 0.25 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 2.4525 \, \text{N} \]

### Step 3: Determine the component of the gravitational force acting down the incline
The component of the weight acting down the incline (parallel to the incline) can be calculated using:
\[ F_{\text{gravity}} = W \cdot \sin(\theta) \]

Calculating this component:
\[ F_{\text{gravity}} = 2.4525 \, \text{N} \cdot \sin(37°) \]
Using the value \( \sin(37°) \approx 0.6 \):
\[ F_{\text{gravity}} = 2.4525 \, \text{N} \cdot 0.6 = 1.4715 \, \text{N} \]

### Step 4: Understand the condition of uniform speed
Since the block is sliding down with uniform speed, the net force acting on it is zero. This implies that the frictional force (F_friction) acting against the motion is equal to the component of the gravitational force acting down the incline:
\[ F_{\text{friction}} = F_{\text{gravity}} = 1.4715 \, \text{N} \]

### Step 5: Calculate the work done against friction
The work done against friction (W_friction) can be calculated using the formula:
\[ W_{\text{friction}} = F_{\text{friction}} \cdot d \cdot \cos(180°) \]
Since the friction force acts opposite to the direction of displacement, we use \( \cos(180°) = -1 \):
\[ W_{\text{friction}} = 1.4715 \, \text{N} \cdot 1.0 \, \text{m} \cdot (-1) \]
\[ W_{\text{friction}} = -1.4715 \, \text{J} \]

### Final Answer
The work done against friction as the block slides through 1.0 m is approximately:
\[ W_{\text{friction}} \approx -1.47 \, \text{J} \]

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