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

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

The displacement $s$ of the block along the incline can be calculated as:

$W = Fs \Rightarrow 40 = 20s \Rightarrow s = 2 \text{ m}$

Using the kinematic equation $s = ut + \dfrac{1}{2}at^2$ with initial velocity $u = 0$ and time $t = 1 \text{ s}$:

$2 = 0 + \dfrac{1}{2}a(1)^2 \Rightarrow a = 4 \text{ m/s}^2$

The velocity of the block at $t = 1 \text{ s}$ is:

$v = u + at = 0 + 4(1) = 4 \text{ m/s}$

The kinetic energy of the block at this instant is:

$K = \dfrac{1}{2}mv^2 = \dfrac{1}{2}(2.0)(4)^2 = 16 \text{ J}$

Alternatively, using the work-energy theorem:

The component of gravitational force acting downwards along the incline is mg \sin 37^{}^{{}^{\circ}} = 2.0 \times 10 \times \dfrac{3}{5} = 12 \text{ N}.

Work done by gravity is W_g = -mg \sin 37^{}^{{}^{\circ}} \times s = -12 \times 2 = -24 \text{ J}.

The net work done on the block is $W_{\text{net}} = W + W_g = 40 - 24 = 16 \text{ J}$.

By the work-energy theorem, $\Delta K = W_{\text{net}} \Rightarrow K - 0 = 16 \text{ J}$.