A block of mass 4 kg slides down an inclined surface having an...

Let the initial acceleration of the block be $a_1$.

Using the equation of motion $s = ut + \dfrac{1}{2}a_1 t^2$:

$8 = 0 + \dfrac{1}{2}a_1 (2)^2$

$a_1 = 4 \text{ m/s}^2$

The net force acting on the block initially is:

$F_{net} = ma_1 = 4 \times 4 = 16 \text{ N}$

When an additional force of $4 \text{ N}$ is applied to push the block down the incline, the new net force becomes:

$F'_{net} = F_{net} + 4 = 16 + 4 = 20 \text{ N}$

The new acceleration of the block is:

$a_2 = \dfrac{F'_{net}}{m} = \dfrac{20}{4} = 5 \text{ m/s}^2$

The distance covered by the block in the first two seconds with this new acceleration is:

$s' = ut + \dfrac{1}{2}a_2 t^2$

$s' = 0 + \dfrac{1}{2}(5)(2)^2 = 10 \text{ m}$