91Ó°ÊÓ

Mass of the block is$m=250\mathrm{kg}$

Spring constant is$k=2.5\mathrm{N}/\mathrm{cm}=250\mathrm{N}/\mathrm{m}$

Compressed distance is$d=12\mathrm{cm}=0.12\mathrm{m}$

This problem deals with the work-energy theorem which states that the net work done by the forces applied on object is equal to the change in its kinetic energy. To solve this problem, use the work-energy theorem and the equation relating work done and force.

Formulae:

$$\begin{gathered} W=Fx\cos \mathrm{ϕ} \\ W=∆K \end{gathered}$$

Where,

F is force, xis displacement,$∆K$ is change in kinetic energy and Wis the work done.

Hence, the work done by gravitational force is $W_g=0.29\mathrm{J}$

Kinetic energy of the spring is

$$\begin{gathered} K=\frac{1}{2}k{d}^2 \\ {W}_s=∆K \\ =K_i-K_f \end{gathered}$$

As the initial position of the spring is mean position,

role="math" localid="1657176614801" $$\begin{gathered} W_s=-K_f \\ {W}_s=-\frac{1}{2}k{d}^2\left(\mathrm{ii}\right) \\ {W}_s=-\frac{1}{2}×250×{(0.12)}^2{W}_s=-1.8\mathrm{J} \end{gathered}$$

Hence, the work done by the spring is$W_s=-1.8\mathrm{J}$

$W_{net}=∆K=K_i-K_f$

Astheinitial position of the spring is mean position,$K_i=0$,

$$\begin{gathered} W_{net}=-K_f \\ {W}_s+W_g=-\frac{1}{2}m{v}_f^2 \\ -1.8+0.294=-\frac{1}{2}×250×v_f^2 \\ {v}_f^2=\sqrt{\left(\frac{2×1.506}{0.25}\right)} \\ {v}_f=3.471\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{f}}=3.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the object before it hit the spring$v=3.5\mathrm{m}/\mathrm{s}$.

$$\begin{gathered} W_{net}=-Kf \\ {W}_s+W_g=-\frac{1}{2}m{v}_{{}^f}^2 \\ U\sin g{v}_f=2v_f\mathrm{and}\mathrm{equation}(iand2, \\ -\frac{1}{2}k{d}^2+mgd=-\frac{1}{2}m{\left(2v_f\right)}^2\mathrm{Solving}\mathrm{and}\mathrm{rearranging}, \\ {d}^2-\frac{2mgd}{k}-\frac{m{v}_f^2}{k}=0 \\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{quardratic}\mathrm{equation}\mathrm{with}, \\ a=1,b=-\frac{2mg}{k},c=-\frac{m{v}_f^2}{k} \\ \mathrm{As}\mathrm{the}\mathrm{negative}\mathrm{value}\mathrm{will}\mathrm{give}\mathrm{a}\mathrm{negative}\mathrm{root},\mathrm{consider}\mathrm{only}\mathrm{the}\mathrm{positive}\mathrm{factor}, \\ d=\frac{-b+\sqrt{b^2-4ac}}{2a} \\ d=\frac{-\left(-{\displaystyle \frac{2mg}{k}}\right)+\sqrt{{\left(-{\displaystyle \frac{2mg}{k}}\right)}^2-4\left(-{\displaystyle \frac{\mathit{m}{\mathit{v}}_{\mathit{f}}^{\mathit{2}}}{\mathit{k}}}\right)}}{2\mathrm{a}} \\ d=\frac{\left({\displaystyle \frac{2mg}{k}}\right)+{\displaystyle \frac{2}{k}}\sqrt{\left({\displaystyle \frac{4m^2{g}^2}{k^2}}\right)+4{\displaystyle \frac{m{v}_f^2}{k}}}}{2} \\ d=\frac{\left({\displaystyle \frac{2mg}{k}}\right)+{\displaystyle \frac{2}{k}}\sqrt{m^2{k}^2+m{v}_f^{2}k}}{2} \\ d=\frac{mg+\sqrt{m^2{g}^2+m{v}_f^{2}k}}{k} \\ d=\frac{0.25×9.8+\sqrt{{\left(0.25\right)}^2{(9.8)}^2+0.25{(3.471)}^{2}250}}{250} \\ \mathrm{By}\mathrm{solving}\mathrm{the}\mathrm{equation}, \\ d=0.23\mathrm{m} \end{gathered}$$

Hence, if$v^{\text{'}}=2v$, then distance of compression is $d=0.23\mathrm{m}$.