A $$0.4$$ kg mass takes $$8$$ s to reach ground when dropped from a certain height $$P$$ above surface of earth. The loss of potential energy in the last second of fall is ______ J. [Take $$g = 10$$ m s$$^{-2}$$]

Find the loss of potential energy in the last second of fall for a 0.4 kg mass dropped from height P, taking 8 seconds to reach ground.

$$h_8 = \frac{1}{2}g t^2 = \frac{1}{2} \times 10 \times 64 = 320 \text{ m}$$

$$h_7 = \frac{1}{2} \times 10 \times 49 = 245 \text{ m}$$

$$\Delta h = h_8 - h_7 = 320 - 245 = 75 \text{ m}$$

$$\Delta PE = mg\Delta h = 0.4 \times 10 \times 75 = 300 \text{ J}$$

The answer is $$\boxed{300}$$.