A particle of mass 1kg travels in a straight line with a velocity $v=a{x}^{\frac{1}{2}}$ , a=5.  What is work done by a net force during its replacement from x=0 to $x=\text{2}{\text{m}}^{\text{2}}$

Given mass of particle $m=1\text{kg}$

Velocity $v=a{x}^{\frac{1}{2}}$

Where  a = constant = 5

At $x=0,v=a{x}^{\frac{1}{2}}=0$

At $x=2,v=a{x}^{\frac{1}{2}}=5\times 2^{\frac{1}{2}}$

According to work energy theorem. Work done = change in kinetic energy

$w=\Delta KE=\frac{1}{2}m{v}^2-0=\frac{1}{2}\times 1$

$W=\frac{1}{2}m{v}^2-0=\frac{1}{2}\times {\left(5\times 2^{\frac{1}{2}}\right)}^2$

$=\frac{1}{2}\times 25\times 2=25\text{J}$