To solve the problem step by step, we will follow the outlined process below:
### Step 1: Identify the Given Data
- Mass of the block, \( m = 2 \, \text{kg} \)
- Coefficient of static friction, \( \mu_s = 0.6 \)
- Coefficient of kinetic friction, \( \mu_k = 0.2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
### Step 2: Calculate the Normal Force
The normal force \( N \) acting on the block is equal to its weight since it is on a horizontal surface:
\[
N = mg = 2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N}
\]
### Step 3: Calculate the Maximum Static Friction
The maximum static friction force \( f_s \) that needs to be overcome to start the motion is given by:
\[
f_s = \mu_s \cdot N = \mu_s \cdot mg
\]
Substituting the values:
\[
f_s = 0.6 \cdot 20 \, \text{N} = 12 \, \text{N}
\]
### Step 4: Determine the Minimum Force Required to Start Motion
The minimum horizontal force \( F \) required to start the motion is equal to the maximum static friction:
\[
F = f_s = 12 \, \text{N}
\]
### Step 5: Calculate the Kinetic Friction Force
Once the block starts moving, the kinetic friction force \( f_k \) acting on it is given by:
\[
f_k = \mu_k \cdot N = \mu_k \cdot mg
\]
Substituting the values:
\[
f_k = 0.2 \cdot 20 \, \text{N} = 4 \, \text{N}
\]
### Step 6: Calculate the Net Force Acting on the Block
The net force \( F_{net} \) acting on the block after it starts moving is:
\[
F_{net} = F - f_k = 12 \, \text{N} - 4 \, \text{N} = 8 \, \text{N}
\]
### Step 7: Calculate the Acceleration of the Block
Using Newton's second law, the acceleration \( a \) of the block can be calculated as:
\[
a = \frac{F_{net}}{m} = \frac{8 \, \text{N}}{2 \, \text{kg}} = 4 \, \text{m/s}^2
\]
### Step 8: Calculate the Velocity After 2 Seconds
Using the equation of motion, the final velocity \( v \) after time \( t = 2 \, \text{s} \) can be calculated using:
\[
v = u + at
\]
where \( u \) (initial velocity) is 0 (since the block starts from rest):
\[
v = 0 + (4 \, \text{m/s}^2 \cdot 2 \, \text{s}) = 8 \, \text{m/s}
\]
### Final Answer
The velocity acquired by the body at the end of the 2nd second is:
\[
\boxed{8 \, \text{m/s}}
\]
