A block of mass `10 kg` slides down a rough slope which is inclined at `45^(0)` to the horizontal. The coeffficient of sliding friction is `0.30`. When the block has to slide `5m`, the work done on the block by the force of friction is nearly

Options

A. `115 J`
B. `-75 sqrt(2) J`
C. `321.4 J`
D. `-321.4 J`

Explanation

To solve the problem step by step, we will calculate the work done by the force of friction on a block sliding down a rough slope.

### Step 1: Identify the forces acting on the block
The block experiences three main forces:
1. Gravitational force (weight) acting downwards: \( F_g = mg \)
2. Normal force acting perpendicular to the slope: \( N \)
3. Frictional force acting opposite to the direction of motion: \( f \)

### Step 2: Calculate the gravitational force
Given:
- Mass of the block, \( m = 10 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

The gravitational force is:
\[ F_g = mg = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N} \]

### Step 3: Resolve the gravitational force into components
The slope is inclined at \( 45^\circ \). We resolve the gravitational force into two components:
- Parallel to the slope: \( F_{\parallel} = mg \sin(45^\circ) \)
- Perpendicular to the slope: \( F_{\perpendicular} = mg \cos(45^\circ) \)

Calculating these:
\[ F_{\parallel} = 100 \, \text{N} \times \sin(45^\circ) = 100 \, \text{N} \times \frac{1}{\sqrt{2}} = 50\sqrt{2} \, \text{N} \]
\[ F_{\perpendicular} = 100 \, \text{N} \times \cos(45^\circ) = 100 \, \text{N} \times \frac{1}{\sqrt{2}} = 50\sqrt{2} \, \text{N} \]

### Step 4: Calculate the normal force
The normal force \( N \) is equal to the perpendicular component of the gravitational force:
\[ N = F_{\perpendicular} = 50\sqrt{2} \, \text{N} \]

### Step 5: Calculate the frictional force
The frictional force \( f \) is given by:
\[ f = \mu N \]
Where \( \mu = 0.3 \) (coefficient of friction). Thus:
\[ f = 0.3 \times 50\sqrt{2} = 15\sqrt{2} \, \text{N} \]

### Step 6: Calculate the work done by the frictional force
The work done by the frictional force \( W \) is given by:
\[ W = f \cdot d \cdot \cos(\phi) \]
Where:
- \( d = 5 \, \text{m} \) (distance slid)
- \( \phi = 180^\circ \) (angle between friction and displacement)

Since \( \cos(180^\circ) = -1 \):
\[ W = 15\sqrt{2} \, \text{N} \times 5 \, \text{m} \times (-1) = -75\sqrt{2} \, \text{J} \]

### Final Answer
The work done on the block by the force of friction is approximately:
\[ W \approx -75\sqrt{2} \, \text{J} \]