A block of mass 2kg initially at rest moves under the action of an applied horizontal force of 6 N on a rough horizontal surface. The coefficient of friction between the block and surface is 0.1. the work done by friction in 10 s is

Options

A. 1) 200J
B. 2) `-200 J`
C. 3) 600 J
D. 4) `-600 J`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the forces acting on the block
- The block has a mass \( m = 2 \, \text{kg} \).
- An applied horizontal force \( F_{\text{applied}} = 6 \, \text{N} \).
- The coefficient of friction \( \mu = 0.1 \).
- The gravitational acceleration \( g = 10 \, \text{m/s}^2 \).

### Step 2: Calculate the normal force
Since the block is on a horizontal surface, the normal force \( N \) is equal to the weight of the block:
\[ N = mg = 2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N} \]

### Step 3: Calculate the frictional force
The frictional force \( F_{\text{friction}} \) can be calculated using the formula:
\[ F_{\text{friction}} = \mu N = 0.1 \times 20 \, \text{N} = 2 \, \text{N} \]

### Step 4: Calculate the net force acting on the block
The net force \( F_{\text{net}} \) acting on the block is given by:
\[ F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 6 \, \text{N} - 2 \, \text{N} = 4 \, \text{N} \]

### Step 5: Calculate the acceleration of the block
Using Newton's second law, we can find the acceleration \( a \):
\[ F_{\text{net}} = ma \implies 4 \, \text{N} = 2 \, \text{kg} \times a \implies a = \frac{4 \, \text{N}}{2 \, \text{kg}} = 2 \, \text{m/s}^2 \]

### Step 6: Calculate the displacement of the block in 10 seconds
Using the formula for displacement \( s \):
\[ s = \frac{1}{2} a t^2 = \frac{1}{2} \times 2 \, \text{m/s}^2 \times (10 \, \text{s})^2 = \frac{1}{2} \times 2 \times 100 = 100 \, \text{m} \]

### Step 7: Calculate the work done by friction
The work done by the frictional force \( W_{\text{friction}} \) is given by:
\[ W_{\text{friction}} = F_{\text{friction}} \times s \times \cos(\theta) \]
Since the frictional force acts in the opposite direction to the displacement, \( \theta = 180^\circ \) and \( \cos(180^\circ) = -1 \):
\[ W_{\text{friction}} = 2 \, \text{N} \times 100 \, \text{m} \times (-1) = -200 \, \text{J} \]

### Final Answer
The work done by friction in 10 seconds is \( \boxed{-200 \, \text{J}} \).

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