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

The displacement of the block in the first second is given by the equation of motion:

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

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 forces acting on the block parallel to the inclined plane are the applied force $F$, the component of gravity $mg \sin \theta$, and the frictional force $f$ acting opposite to the motion. The equation of motion is:

$F + mg \sin \theta - f = ma$

Substituting $F = 20 \text{ N}$, $m = 2.0 \text{ kg}$, $g = 10 \text{ m/s}^2$, \theta = 37^{}^{{}^{\circ}}, and $a = 10 \text{ m/s}^2$:

20 + (2.0)(10) \sin 37^{}^{{}^{\circ}} - f = (2.0)(10)

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

$20 + 20 \left(\dfrac{3}{5}\right) - f = 20$

$20 + 12 - f = 20$

$f = 12 \text{ N}$

The work done by the frictional force is the product of the frictional force and the displacement. Since friction acts in the direction opposite to the displacement:

$W_f = -f \times s$

$W_f = -12 \times 5 = -60 \text{ J}$