A block of mass 1 kg slides down on a rough inclined plane of inclination `60^@` starting from its top. If the coefficient of kinetic friction is 0.5 and length of the plane is 1 m, then work done against friction is `("Take g" = 9.8 m//s^2)`

Options

A. 9.82 J
B. 4.94 J
C. 2.45J
D. 1.96 J

Explanation

To solve the problem of calculating the work done against friction for a block sliding down a rough inclined plane, we can follow these steps:

### Step 1: Identify the given values
- Mass of the block (m) = 1 kg
- Coefficient of kinetic friction (μ_k) = 0.5
- Angle of inclination (θ) = 60 degrees
- Length of the inclined plane (S) = 1 m
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Calculate the normal force (N)
The normal force on an inclined plane can be calculated using the formula:
\[ N = m \cdot g \cdot \cos(\theta) \]

Substituting the values:
\[ N = 1 \cdot 9.8 \cdot \cos(60^\circ) \]
Since \(\cos(60^\circ) = \frac{1}{2}\):
\[ N = 1 \cdot 9.8 \cdot \frac{1}{2} = 4.9 \, \text{N} \]

### Step 3: Calculate the frictional force (F_friction)
The frictional force can be calculated using the formula:
\[ F_{\text{friction}} = \mu_k \cdot N \]

Substituting the values:
\[ F_{\text{friction}} = 0.5 \cdot 4.9 = 2.45 \, \text{N} \]

### Step 4: Calculate the work done against friction (W)
The work done against friction can be calculated using the formula:
\[ W = F_{\text{friction}} \cdot S \]

Substituting the values:
\[ W = 2.45 \cdot 1 = 2.45 \, \text{J} \]

### Conclusion
The work done against friction as the block slides down the inclined plane is **2.45 Joules**.

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