A body of mass 5 kg is moving with a momentum of 10 kg m s$$^{-1}$$. Now a force of 2 N acts on the body in the direction of its motion for 5 s. The increase in the Kinetic energy of the body is _______ J.

We have a body of mass $$m = 5$$ kg with initial momentum $$p_i = 10$$ kg m/s, acted upon by a force $$F = 2$$ N for $$t = 5$$ s.

The initial velocity is $$u = \frac{p_i}{m} = \frac{10}{5} = 2$$ m/s, so the initial kinetic energy is:

$$K_i = \frac{1}{2}mu^2 = \frac{1}{2}(5)(4) = 10$$ J

Now, using the impulse-momentum theorem to find the final momentum:

$$p_f = p_i + Ft = 10 + 2 \times 5 = 20 \text{ kg m/s}$$

So the final velocity is $$v = \frac{p_f}{m} = \frac{20}{5} = 4$$ m/s, and the final kinetic energy is:

$$K_f = \frac{1}{2}mv^2 = \frac{1}{2}(5)(16) = 40$$ J

Hence, the increase in kinetic energy is:

$$\Delta K = K_f - K_i = 40 - 10 = 30 \text{ J}$$

So, the answer is $$30$$ J.