A body of mass 10 kg is moving on a horizonal surface be applying a force of 10 N in the forward direction. The body moves with a constant velocity of `0.2ms^(-1)`. Work done by the force of friction in the first 10 seconds is

Options

A. `-20 J`
B. 10 J
C. 20 J
D. `-5J`

Answer

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Understand the Given Information
We have a body of mass \( m = 10 \, \text{kg} \) moving on a horizontal surface with a constant velocity of \( v = 0.2 \, \text{m/s} \). A force of \( F = 10 \, \text{N} \) is applied in the forward direction.

### Step 2: Identify the Forces Acting on the Body
Since the body is moving with a constant velocity, according to Newton's first law of motion, the net force acting on the body must be zero. This means that the applied force (10 N) is balanced by the frictional force acting in the opposite direction.

### Step 3: Calculate the Frictional Force
Since the body is moving at constant velocity, the frictional force \( F_f \) must equal the applied force:
\[ F_f = F = 10 \, \text{N} \]

### Step 4: Calculate the Displacement in 10 Seconds
To find the work done by the frictional force, we first need to calculate the displacement of the body over the first 10 seconds. Using the formula for displacement:
\[ \text{Displacement} = \text{Velocity} \times \text{Time} \]
Substituting the known values:
\[ \text{Displacement} = 0.2 \, \text{m/s} \times 10 \, \text{s} = 2 \, \text{m} \]

### Step 5: Calculate the Work Done by the Frictional Force
Work done \( W \) by the frictional force can be calculated using the formula:
\[ W = F_f \times d \times \cos(\theta) \]
Where:
- \( F_f = 10 \, \text{N} \) (frictional force)
- \( d = 2 \, \text{m} \) (displacement)
- \( \theta = 180^\circ \) (the angle between the frictional force and displacement is 180 degrees, as they act in opposite directions)

Since \( \cos(180^\circ) = -1 \), we have:
\[ W = 10 \, \text{N} \times 2 \, \text{m} \times (-1) = -20 \, \text{J} \]

### Final Answer
The work done by the force of friction in the first 10 seconds is:
\[ \boxed{-20 \, \text{J}} \]