A body of mass 5 kg is moving with a momentum of 10 kg-m/s. A force of 0.2 N acts on it in the direction of motion of the body for 10 seconds. The increase in its kinetic energy is

The correct option is 4.4 J.

Let's break down the solution step-by-step to find the increase in the kinetic energy of the body.

Step 1: Identify the given values from the problem statement:
- Mass of the body, $m=5\text{kg}$
- Initial momentum, $p_i=10\text{kg·m/s}$
- Applied force, $F=0.2\text{N}$
- Time duration, $t=10\text{s}$

Step 2: Find the initial velocity ($v_i$) of the body:
We know that momentum is the product of mass and velocity:
$p=mv$
So, the initial velocity is:
$v_i=\frac{p_i}{m}=\frac{10}{5}=2\text{m/s}$

Step 3: Calculate the acceleration ($a$) produced by the force:
Using Newton's second law of motion ($F=ma$):
$a=\frac{F}{m}=\frac{0.2}{5}=0.04{\text{m/s}}^2$

Step 4: Determine the final velocity ($v_f$) after 10 seconds:
Using the first equation of motion:
$v_f=v_i+at$
$v_f=2+(0.04\times 10)=2+0.4=2.4\text{m/s}$

Step 5: Calculate the increase in kinetic energy ($\mathrm{\Delta K.E.}$):
The increase in kinetic energy is the difference between the final and initial kinetic energies:
$\mathrm{\Delta K.E.}=\frac{1}{2}m(v_f^2-v_i^2)$
Substitute the values into the equation:
$\mathrm{\Delta K.E.}=\frac{1}{2}\times 5\times ({2.4}^2-2^2)$
$\mathrm{\Delta K.E.}=2.5\times (5.76-4)$
$\mathrm{\Delta K.E.}=2.5\times 1.76=4.4\text{J}$

Therefore, the increase in the kinetic energy of the body is 4.4 J.