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

Given the work done by the applied force $W_F = 40 \text{ J}$ and the applied force $F = 20 \text{ N}$.

The displacement of the block along the incline is given by:

$s = \dfrac{W_F}{F} = \dfrac{40}{20} = 2 \text{ m}$

The block moves $2 \text{ m}$ up the incline. The vertical displacement of the block is h = s \sin 37^{}^{{}^{\circ}}.

The work done by the force of gravity is negative since the displacement is upwards while gravity acts downwards:

W_g = -mgh = -mgs \sin 37^{}^{{}^{\circ}}

Substituting the given values ($m = 2.0 \text{ kg}$, $g = 10 \text{ m/s}^2$, \sin 37^{}^{{}^{\circ}} = \dfrac{3}{5}):

$W_g = -2.0 \times 10 \times 2 \times \dfrac{3}{5}$

$W_g = -12 \times 2 = -24 \text{ J}$