A body of mass 0.5 kg travels in a straight line with velocity v = a x¹.⁵ where a = 5 m⁻¹.⁵s⁻¹. The work done by the net force during its displacement from x = 0 to x = 2 m is

To find the work done by the net force, we can apply the work-energy theorem, which states that the net work done on a body is equal to the change in its kinetic energy:

$W=\Delta K=K_{\text{f}}-K_{\text{i}}$

where $K_{\text{i}}$ is the initial kinetic energy and $K_{\text{f}}$ is the final kinetic energy.

The kinetic energy $K$ of a body of mass $m$ moving with velocity $v$ is given by:

$K=\frac{1}{2}m{v}^2$

We are given the relation for velocity as a function of displacement:
$v=a{x}^{1.5}$
with $a=5{\text{m}}^{-0.5}{\text{s}}^{-1}$ and mass $m=0.5\text{kg}$.

Let's calculate the initial velocity $v_{\text{i}}$ at the starting position $x=0\text{m}$:

$v_{\text{i}}=5\times {\left(0\right)}^{1.5}=0\text{m/s}$

Therefore, the initial kinetic energy is:

$K_{\text{i}}=\frac{1}{2}m{v}_{\text{i}}^2=0\text{J}$

Next, let's calculate the final velocity $v_{\text{f}}$ at the final position $x=2\text{m}$:

$v_{\text{f}}=5\times 2^{1.5}=5\times 2^{3/2}=5\times 2\sqrt{2}=10\sqrt{2}\text{m/s}$

Now, we calculate the final kinetic energy $K_{\text{f}}$:

$K_{\text{f}}=\frac{1}{2}m{v}_{\text{f}}^2=\frac{1}{2}\times 0.5\times {\left(10\sqrt{2}\right)}^2$

Evaluating this expression:

$K_{\text{f}}=0.25\times (100\times 2)=0.25\times 200=50\text{J}$

Thus, the work done by the net force is:

$W=K_{\text{f}}-K_{\text{i}}=50\text{J}-0\text{J}=50\text{J}$

The correct option is 50 J.