A block of mass 2.0 kg initially at rest on an inclined plane of...

The work done by the applied force is $W = Fs$, where $F = 20 \text{ N}$ and $s$ is the displacement of the block along the incline in $1 \text{ s}$.

To maximize the work done, the displacement $s$ must be maximum, which requires the acceleration $a$ of the block to be maximum.

The forces acting on the block along the incline are the applied force $F$ upwards, the gravitational component mg \sin 37^{}^{{}^{\circ}} downwards, and the frictional force $f$ downwards.

The equation of motion is F - mg \sin 37^{}^{{}^{\circ}} - f = ma.

For maximum acceleration, the frictional force must be zero ($f = 0$).

Substituting the given values:

20 - 2.0 \times 10 \times \sin 37^{}^{{}^{\circ}} = 2.0 \times a

$20 - 20 \times \dfrac{3}{5} = 2.0 a$

$20 - 12 = 2.0 a \Rightarrow a = 4 \text{ m/s}^2$

The maximum displacement in $t = 1 \text{ s}$ starting from rest is:

$s = \dfrac{1}{2} a t^2 = \dfrac{1}{2} \times 4 \times (1)^2 = 2 \text{ m}$

The maximum possible work done by the applied force is:

$W = Fs = 20 \times 2 = 40 \text{ J}$