A block of mass 2.0 kg is pushed down an inclined plane of...

The distance traveled by the block in the first second is given by the kinematic equation:

$s = ut + \dfrac{1}{2}at^2$

Since the block starts from rest, $u = 0$. Substituting the given values $a = 10 \text{ m/s}^2$ and $t = 1 \text{ s}$:

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

The weight of the block acts vertically downwards, and its component parallel to the inclined plane is $mg \sin \theta$. The work done by the weight of the block is the product of this parallel component and the displacement along the incline:

$W = (mg \sin \theta) \times s$

Substituting $m = 2.0 \text{ kg}$, $g = 10 \text{ m/s}^2$, \theta = 37^{}^{{}^{\circ}}, and $s = 5 \text{ m}$:

W = (2.0 \times 10 \times \sin 37^{}^{{}^{\circ}}) \times 5

Using \sin 37^{}^{{}^{\circ}} = \dfrac{3}{5}:

$W = 20 \times \dfrac{3}{5} \times 5 = 60 \text{ J}$