A

Given:
Mass of the body, m = 5 kg
Force applied, F = 20 N
Time, t = 10 s

Step 1: Calculate the acceleration of the body using Newton's second law.
According to Newton's second law:
$F = m \cdot a$
Rearranging gives us:
$$a = \frac{F}{m} = \frac{20 \text{ N}}{5 \text{ kg}} = 4 \text{ m/s}^2$$

Step 2: Calculate the distance moved by the body in 10 seconds.
Since the initial velocity (u) is 0 (the body is initially at rest), we can use the equation of motion:
$$s = ut + \frac{1}{2} a t^2$$
Substituting the known values:
$$s = 0 + \frac{1}{2} \cdot 4 \cdot (10)^2 = 200 \text{ m}$$

Step 3: Calculate the work done by the force.
Work done (W) is given by the formula:
$$W = F \cdot d$$
Here, d is the distance moved.
$$W = 20 \text{ N} \cdot 200 \text{ m} = 4000 \text{ J}$$

Step 4: Calculate the final kinetic energy of the body.
The final velocity (v) after 10 seconds can be calculated using:
$$v = u + at = 0 + 4 \cdot 10 = 40 \text{ m/s}$$
Now, the kinetic energy (KE) is given by:
$$KE = \frac{1}{2} mv^2 = \frac{1}{2} \cdot 5 \cdot (40)^2 = \frac{1}{2} \cdot 5 \cdot 1600 = 4000 \text{ J}$$

Step 5: Prove that work done equals the change in kinetic energy.
The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.
Since the body started from rest, the initial kinetic energy is 0.
Thus, the change in kinetic energy is:
$$\Delta KE = KE_{final} - KE_{initial} = 4000 \text{ J} - 0 = 4000 \text{ J}$$

Conclusion:
Work done by the force (4000 J) equals the change in kinetic energy (4000 J). Therefore, the calculation and the theorem hold true.