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

Options

A. 2.45 J
B. 4.9 J
C. 9.8 J
D. 19.6 J

Explanation

To find the work done against friction when a block of mass 1 kg slides down a rough inclined plane of inclination 60 degrees, we can follow these steps:

### Step 1: Identify the Forces Acting on the Block
The forces acting on the block are:
- Weight (mg) acting vertically downward.
- Normal force (N) acting perpendicular to the inclined plane.
- Frictional force (fk) acting opposite to the direction of motion.

### Step 2: Calculate the Normal Force (N)
The normal force can be calculated using the formula:
\[ N = mg \cos(\theta) \]
where:
- \( m = 1 \, \text{kg} \) (mass of the block)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( \theta = 60^\circ \)

Substituting the values:
\[ N = 1 \times 9.8 \times \cos(60^\circ) \]
Since \( \cos(60^\circ) = 0.5 \):
\[ N = 1 \times 9.8 \times 0.5 = 4.9 \, \text{N} \]

### Step 3: Calculate the Frictional Force (fk)
The frictional force can be calculated using the formula:
\[ f_k = \mu N \]
where:
- \( \mu = 0.5 \) (coefficient of kinetic friction)

Substituting the values:
\[ f_k = 0.5 \times 4.9 = 2.45 \, \text{N} \]

### Step 4: Calculate the Work Done Against Friction (W)
The work done against friction is given by:
\[ W = -f_k \cdot d \]
where:
- \( d = 2 \, \text{m} \) (length of the inclined plane)

Substituting the values:
\[ W = -2.45 \times 2 = -4.9 \, \text{J} \]

### Final Answer
The work done against friction is:
\[ W = -4.9 \, \text{J} \]

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