Physics - W.E.P. Question with Solution | TestHub

Given : ${\mathrm{k}}_{\mathrm{A}}=300\mathrm{N}/\mathrm{m},{\mathrm{k}}_{\mathrm{B}}=400\mathrm{N}/\mathrm{m}$ Let when the combination of springs is compressed by force F. Spring A is compressed by $x$. Therefore compression in spring $\mathrm{B}$ $$\begin{array}{rl}& x_B=(8.75-x)\mathrm{cm}\\ & \mathrm{F}=300\times \mathrm{x}=400(8.75-\mathrm{x})\end{array}$$ Solving we get, $x=5\mathrm{cm}$ $${\mathrm{x}}_{\mathrm{B}}=8.75-5=3.75\mathrm{cm}$$ $\frac{{\mathrm{E}}_{\mathrm{A}}}{{\mathrm{E}}_{\mathrm{B}}}=\frac{\frac{1}{2}{\mathrm{k}}_{\mathrm{A}}{\left({\mathrm{x}}_{\mathrm{A}}\right)}^2}{\frac{1}{2}{\mathrm{k}}_{\mathrm{B}}{\left({\mathrm{x}}_{\mathrm{B}}\right)}^2}=\frac{300\times (5{)}^2}{400\times (3.75{)}^2}=\frac{4}{3}$