Question

A particle of mass m moves on a straight line with its velocity varying with the distance travelled, according to the equation $\nu =a\sqrt{x}$, where a is a constant. Find the total work done by all the forces during a displacement from $x=0\mathrm{to}x-d$.

Answer 1

Given,
$$\begin{gathered} \nu =a\sqrt{x}\left(\mathrm{uniformly}\mathrm{accelerated}\mathrm{motion}\right) \\ \text{D}\mathrm{isplacement},s=d-0=d \\ \text{P}\mathrm{utting}x=0,\text{we get}{\nu }_1=0 \\ \text{P}\mathrm{utting}x=d,\text{we get}{\nu }_2=a\sqrt{d} \\ \alpha =\frac{{\nu }_2^2-{\nu }_1^2}{2s}=\frac{a^{2}d}{2d}=\frac{a^2}{2} \end{gathered}$$
$$\begin{gathered} \mathrm{Force},F=m\alpha =\frac{m{a}^2}{2} \\ \mathrm{Work}\mathrm{done},\mathrm{W}=Fs\cos \mathrm{\theta } \\ =\frac{m{a}^2}{2}\times d=\frac{m{a}^{2}d}{2} \end{gathered}$$