1. Given data

• Mass: $m = 0.4\,\text{kg}$
• Total time of fall: $8\,\text{s}$
• Acceleration due to gravity: $g = 10\,\text{m/s}^2$

We need the loss of potential energy in the last second of fall, i.e. between $t=7\,\text{s}$ and $t=8\,\text{s}$.

2. Distance fallen in $t$ seconds

Since the body is dropped from rest,

$s = \frac{1}{2}gt^2$

So,

• Distance fallen in $8$ s: $s_8 = \frac{1}{2}\cdot 10 \cdot 8^2 = 5\cdot 64 = 320\,\text{m}$

• Distance fallen in $7$ s: $s_7 = \frac{1}{2}\cdot 10 \cdot 7^2 = 5\cdot 49 = 245\,\text{m}$

Therefore, distance fallen in the last second is

$\Delta h = s_8 - s_7 = 320 - 245 = 75\,\text{m}$

3. Loss of potential energy in the last second

Loss of gravitational potential energy is

$\Delta U = mg\Delta h$

Substitute the values:

$\Delta U = 0.4 \cdot 10 \cdot 75$

$\Delta U = 4 \cdot 75 = 300\,\text{J}$

4. Final answer

The loss of potential energy in the last second is

$\boxed{300\,\text{J}}$