Question: A relaxed spring of lengthstands vertically on the floor; its stiffness is. You release a block of mass from rest, with the bottom of the blockabove the floor and straight above the spring. How long is the spring when the block comes momentarily to rest on the compressed spring?

The elastic potential energy is the energy acquired as a result of applying a force to deform an elastic item. The energy is stored until the force is removed, at which point the object reverts to its original shape and resumes its function. During the deformation process, the object could be crushed, stretched, or twisted.

Let is the relaxed length of the spring, and is its final compressed length.

E_I = E_F

No non-conservative forces act on the system so, the initial energy is equal to the final energy. Equate both the energies.

In the initial state there is only gravitational potential energy, and in the final state, both gravitational and elastic potential energies will present.

Take reference at the floor for the gravitational potential energy and write the initial and final energies into the last equation.

$$\begin{gathered} \text{mgh =}\frac{\text{1}}{\text{2}}{\text{k}}_{\text{s}}{\left({\text{l}}_{\text{0}}\text{- l}\right)}^{\text{2}}\text{+ mgl} \\ \text{mgh =}\frac{\text{1}}{\text{2}}{\text{k}}_{\text{s}}\left({\text{l}}_{\text{0}}^{\text{2}}{\text{- 2l}}_{\text{0}}{\text{l +l}}^{\text{2}}\right)\text{+ mgl} \\ \text{0 =}\frac{\text{1}}{\text{2}}{\text{k}}_{\text{s}}{\text{l}}^{\text{2}}\text{+}\left({\text{mg -k}}_{\text{s}}{\text{l}}_{\text{0}}\right)\text{l +}\left(\frac{\text{1}}{\text{2}}{\text{k}}_{\text{s}}{\text{l}}_{\text{0}}^{\text{2}}\text{- mgh}\right) \end{gathered}$$

Substitute , $k_s=1000 \text{N/m}$,m =0.4 kg , $g=9.8{\text{m/s}}^{\text{2}}$,$I_0=0.15\text{m}$and h =0,8 into the obtained equation and solve for .

Take the negative sign and do the further calculation.

$$\begin{gathered} 0=\frac{1}{2}\left(1000\right)l^2+\left(\left(0.4\right)\left(9.8\right)-\left(1000\right)\left(0.15\right)\right)l+\left[\frac{1}{2}\left(1000\right){\left(0.15\right)}^2-\left(0.4\right)\left(9.8\right)\left(0.8\right)\right] \\ 0=500l^2-146l+8.114 \\ l=\frac{146\pm \sqrt{{(-146)}^2-4\left(500\right)(8.114)}}{2\left(500\right)} \\ =\frac{146\pm 71.33}{1000} \end{gathered}$$

I = 0.0747 m

Therefore, the length of spring obtained is 0.0747 m .