Work Energy and Power

Let's use the work-energy theorem, which says:$K\;=W$$\Rightarrow \;\; K\;=F \times s \;\; \ldots (i)$Here, $K$ is kinetic energy, $F$ is force, and $s$ is distance.When a force acts and the body starts from rest, the distance $s$ covered in time $t$ is:$s = \frac{1}{2} a t^2$Since acceleration $a = \frac{F}{m}$ (where $m$ is mass), we get:$s = \frac{1}{2} \times \frac{F}{m} \times t^2$Now, substitute $s$ from (ii) into our equation (i):$K = F \times \frac{1}{2} \times \frac{F}{m} \times t^2$So, $K = \frac{F^2}{2 m} t^2$We are told both bodies $A$ and $B$ get the same kinetic energy ( $K_A = K_B$ ) using the same force $F = 40\,\mathrm{N}$, but their masses and times are different.Set up the equation for both bodies and compare:$\frac{F^2}{2 m_A} t_A^2 = \frac{F^2}{2 m_B} t_B^2$You can cancel out $F^2$ and $\frac{1}{2}$ because they are the same:$\frac{t_A^2}{m_A} = \frac{t_B^2}{m_B}$Rearrange to solve for time ratio:$\frac{t_A^2}{t_B^2} = \frac{m_A}{m_B}$Take the square root for the ratio $t_A: t_B$ :$\frac{t_A}{t_B} = \sqrt{\frac{m_A}{m_B}}$Plug in $m_A = 20\,\mathrm{kg}$ and $m_B = 5\,\mathrm{kg}$ :$\frac{t_A}{t_B} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2$So, the ratio $t_A: t_B = 2: 1$.