A person slowly slides a block of weight 100 N up a smooth incline...

Let the horizontal driving force applied by the person be $F$.

Since the block is moved slowly, its acceleration is zero. The net force along the incline must be zero. Equating the components of the driving force and the weight along the incline:

$F \cos \theta = W \sin \theta$

The work done by the person is the work done by the horizontal force $F$ over the displacement $s$ along the incline. The angle between the horizontal force and the displacement is $\theta$:

$W_{\text{person}} = F s \cos \theta$

Substituting $F \cos \theta = W \sin \theta$ into the work equation:

$W_{\text{person}} = (W \sin \theta) s = W s \sin \theta$

Given $W = 100 \text{ N}$, $s = 2.0 \text{ m}$, and \theta = 37^{}^{{}^{\circ}}:

W_{\text{person}} = 100 \times 2.0 \times \sin 37^{}^{{}^{\circ}}

$W_{\text{person}} = 200 \times \dfrac{3}{5} = 120 \text{ J}$