A block of mass 5 kg slides down a rough inclined surface. The angle of inclination is `45^(@)`. The coefficient of sliding friction is 0.20. When the block slides 10 cm, the work done on the block by force of friction is

Options

A. `-(1)/(sqrt2)J`
B. 1J
C. `-sqrt2J`
D. `-1J`

Explanation

To solve the problem of finding the work done on a block of mass 5 kg sliding down a rough inclined surface at an angle of 45 degrees with a coefficient of sliding friction of 0.20, we can follow these steps:

### Step 1: Identify the given values
- Mass of the block, \( m = 5 \, \text{kg} \)
- Angle of inclination, \( \theta = 45^\circ \)
- Coefficient of friction, \( \mu = 0.20 \)
- Distance slid, \( s = 10 \, \text{cm} = 0.1 \, \text{m} \)

### Step 2: Calculate the gravitational force acting on the block
The gravitational force \( F_g \) acting on the block can be calculated using the formula:
\[ F_g = m \cdot g \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).
\[ F_g = 5 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 50 \, \text{N} \]

### Step 3: Calculate the normal force
The normal force \( N \) acting on the block on the incline can be calculated as:
\[ N = F_g \cdot \cos(\theta) \]
Substituting the values:
\[ N = 50 \, \text{N} \cdot \cos(45^\circ) = 50 \, \text{N} \cdot \frac{1}{\sqrt{2}} \approx 35.36 \, \text{N} \]

### Step 4: Calculate the frictional force
The frictional force \( F_f \) can be calculated using the formula:
\[ F_f = \mu \cdot N \]
Substituting the values:
\[ F_f = 0.20 \cdot 35.36 \, \text{N} \approx 7.07 \, \text{N} \]

### Step 5: Calculate the work done by the frictional force
The work done \( W \) by the frictional force is given by:
\[ W = -F_f \cdot s \]
The negative sign indicates that the work done by friction is in the opposite direction to the displacement.
Substituting the values:
\[ W = -7.07 \, \text{N} \cdot 0.1 \, \text{m} \approx -0.707 \, \text{J} \]

### Final Answer
The work done on the block by the force of friction is approximately:
\[ W \approx -0.707 \, \text{J} \]

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