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 step by step, we will follow these instructions:

### Step 1: Identify the Given Data
- Mass of the block (m) = 250 g = 0.250 kg (since 1 kg = 1000 g)
- Angle of inclination (θ) = 37°
- Displacement (d) = 1.0 m

### Step 2: Calculate the Weight of the Block
The weight (W) of the block can be calculated using the formula:
\[ W = m \cdot g \]
Where \( g \) is the acceleration due to gravity (approximately 9.8 m/s², but we can use 10 m/s² for simplicity).

\[ W = 0.250 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 2.5 \, \text{N} \]

### Step 3: Resolve the Weight into Components
The weight of the block can be resolved into two components:
1. Perpendicular to the incline: \( W_{\perpendicular} = W \cdot \cos(θ) \)
2. Parallel to the incline: \( W_{\parallel} = W \cdot \sin(θ) \)

Using \( θ = 37° \):
- \( \sin(37°) = \frac{3}{5} \)
- \( \cos(37°) = \frac{4}{5} \)

Calculating the components:
- \( W_{\parallel} = 2.5 \, \text{N} \cdot \sin(37°) = 2.5 \, \text{N} \cdot \frac{3}{5} = 1.5 \, \text{N} \)
- \( W_{\perpendicular} = 2.5 \, \text{N} \cdot \cos(37°) = 2.5 \, \text{N} \cdot \frac{4}{5} = 2.0 \, \text{N} \)

### Step 4: Determine the Frictional Force
Since the block is moving with a uniform speed, the net force acting on it along the incline is zero. Therefore, the frictional force (F_friction) must balance the component of weight acting down the incline:
\[ F_{\text{friction}} = W_{\parallel} = 1.5 \, \text{N} \]

### 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 frictional force acts opposite to the direction of displacement, the angle between the force and displacement is 180° (cos(180°) = -1).

Thus,
\[ W_{\text{friction}} = 1.5 \, \text{N} \cdot 1.0 \, \text{m} \cdot (-1) = -1.5 \, \text{J} \]

### Final Answer
The work done against friction as the block slides through 1.0 m is:
\[ \text{Work done against friction} = 1.5 \, \text{J} \]

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