Given data:
- Mass of the body (m) = 2 kg
- Force applied (F) = 2 T
- Coefficient of friction (μ) = 0.2
- Time (t) = 6 sec

Assumptions:
- The force is applied horizontally.
- The body is initially at rest.

Calculating the acceleration:
From Newton's second law of motion, we know that force is equal to mass times acceleration (F = ma). Rearranging the equation, we can find the acceleration as follows:
a = F/m

Given that F = 2 T and m = 2 kg, we can substitute these values into the equation:
a = 2 T / 2 kg

Converting force to SI units:
1 T = 1000 kg * m/s^2
Therefore, 2 T = 2000 kg * m/s^2

Substituting this value into the equation, we get:
a = 2000 kg * m/s^2 / 2 kg
a = 1000 m/s^2

Calculating the frictional force:
The frictional force (Ff) can be calculated using the formula:
Ff = μ * N

The normal force (N) can be calculated as the product of mass and acceleration due to gravity (g):
N = m * g

Given that the mass (m) = 2 kg and acceleration due to gravity (g) = 9.8 m/s^2, we can substitute these values into the equation:
N = 2 kg * 9.8 m/s^2
N = 19.6 N

Substituting the values of μ = 0.2 and N = 19.6 N into the equation, we get:
Ff = 0.2 * 19.6 N
Ff = 3.92 N

Calculating the net force:
The net force acting on the body can be calculated as the difference between the applied force (F) and the frictional force (Ff):
Net force (Fnet) = F - Ff

Substituting the values of F = 2 T and Ff = 3.92 N into the equation, we get:
Fnet = 2000 kg * m/s^2 - 3.92 N

Calculating the final velocity:
Using the equation of motion v = u + at, where u is the initial velocity (which is 0 m/s in this case), and a is the acceleration calculated earlier, we can find the final velocity (v) at time t = 6 sec:
v = u + at
v = 0 m/s + 1000 m/s^2 * 6 sec
v = 6000 m/s

Therefore, the velocity of the body at t = 6 sec is 6000 m/s.

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