91Ó°ÊÓ

The work energy theorem tells that the work done on the body results in a change in its kinetic energy.

$$\begin{gathered} {\mathbf{W}}_{\mathbf{total}}\mathbf{=}{\mathbf{K}}_{\mathbf{2}}\mathbf{-}{\mathbf{K}}_{\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{∆}\mathbf{K} \end{gathered}$$

From the Work-Energy theorem,

$$\begin{gathered} W=K{E}_f-K{E}_I \\ =\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2 \end{gathered}$$

The initial velocity of the object is zero. So, $v_i=0$.

Substitute 0 for $v_i$, and vfor$v_f$ in the above equation.

$W=\frac{1}{2}m{v}^2        ......\left(1\right)$

Substitute 0 for $v_i$, androle="math" localid="1663744649225" $v_1$ for$v_f$ in the above equation.

$W_1=\frac{1}{2}m{v}_1^2$

Here, the work done by the charged particle is doubled.

$$\begin{gathered} 2\mathrm{W}=\frac{1}{2}{\mathrm{mv}}_1^2 \\ 2\left(\frac{1}{2}{\mathrm{mv}}^2\right)=\frac{1}{2}{\mathrm{mv}}_1^2 \\ 2{\mathrm{v}}^2={\mathrm{v}}_1^2 \\ {\mathrm{v}}_1=\sqrt{2{\mathrm{v}}^2} \\ =\sqrt{2}\mathrm{v} \end{gathered}$$

Therefore, the speed of the object after twice the work is $\sqrt{2}v$.

The initial kinetic energy is given by,

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

The final kinetic energy is given by,

$$\begin{gathered} K\text{'}=\frac{1}{2}m{v}_1^2 \\ =\frac{1}{2}m{\left(\sqrt{2}v\right)}^2 \\ =m{v}^2 \\ =2K \end{gathered}$$

Therefore, the kinetic energy is doubled or 2K.