Physics and only Physics

This is a calculus-based approached because I like calculus.

The integral for the work done is given by $\int_{0}^{d} F dx$

Knowing that $F = ma$ , I took the derivative of $v = k\sqrt{x}$ with respect to time to find acceleration, giving me $\frac{d}{dt}v = \frac{d}{dt}k\sqrt{x} \Rightarrow a = \frac{k}{2\sqrt{x}}\frac{dx}{dt} = \frac{k}{2\sqrt{x}}v = \frac{k^2}{2}$

Multiplying acceleration by the mass $m$ and integrating this constant result from $x = 0$ to $x = d$ gave a final answer of $\int_{0}^{d} \frac{mk^2}{2}dx = \boxed{\frac{1}{2}mk^2d}$

We want to find the work done ( $W$ ) where $W=Fx$ . We also know $F=ma$ , and the distance travelled is $x=d$ .

Therefore $W=mad$ . Now we need to find the value of $a$ , which we can do so with $v^2=u^2+2as$ .

We have $s=d$ , $u=k \sqrt{0}=0$ and $v=k \sqrt{d} ~ \Rightarrow ~ k^2 d =2ad ~ \Rightarrow ~ a=\dfrac{k^2}{2}$ .

Therefore $W=\large\color{#20A900}{\boxed{\dfrac{1}{2}m k^2 d }}$

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