91Ó°ÊÓ

The mass of the cab is, m = 1800 kg

Distance between cab and spring is d = 3.7 m

The spring constant is,

$$\begin{gathered} k=0.15MN/m \\ =0.15×{10}^{6}N/m \end{gathered}$$

The constant frictional force is,

$$\begin{gathered} f=4.4kN \\ =4400N \end{gathered}$$

By using thermal energy$\mathbf{∆}{\mathit{E}}_{\mathbf{t}\mathbf{h}}$and computing gravitational potential U , we can find thespeed v of the cab just before it hits the spring. By finding kinetic energy kand computing gravitational potential U , we can find themaximum distance xthat the spring is compressed. Similarly, assuming d' > xand computing gravitational potential U , we can find thedistance that the cab will bounce back at the shift. By assuming, the elevator comes to a final rest at the equilibrium position of the spring, with the spring compressed an amountrole="math" localid="1661400163825" ${\mathit{d}}_{\mathbf{e}\mathbf{q}}$and computing gravitational potential U , we can find theapproximate total distance${\mathit{d}}_{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}}$ that the cab will move before coming to rest.

Formula:

The thermal energy associated is,$∆E_{th}=f.d$

The gravitational potential energy is,$U=mgy$

Also, it is,$U=K+∆E_{th}$

The kinetic energy is, $K=\frac{1}{2}m{v}^2$

The quadratic formula for x is,$x=\frac{\mathrm{Î\P }Â\pm \sqrt{{\mathrm{Î\P }}^2+2kK}}{k}$

Here, the thermal energy associated is,

$∆E_{th}=f.d$

With W = 0 and computing,

$U=mgy$

Set at the top of the spring, we get,

$$\begin{gathered} U_i=K+∆E_{th} \\ mgd=\frac{1}{2}m{v}^2+fd \\ v=\sqrt{2d\left(g-\frac{f}{m}\right)} \\ =\sqrt{2×3.7×\left(9.8-\frac{4400}{1800}\right)} \\ =7.4m/s \end{gathered}$$

Hence, the speed of the cab just before it hits the spring is v = 7.4 m/s

Again with, W = 0 and computing, U = mgy

Therefore, we end up with gravitational potential energy at the bottom point as

Gravitational potential energy = mg(-x)

Therefore, we get,

$K=mg\left(-x\right)+\frac{1}{2}{x}^2+fx$

But

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}×1800×7.4^2 \\ =4.9×{10}^{4}J \end{gathered}$$

And

$$\begin{gathered} \mathrm{Î\P }=mg-f \\ =\left(1800×9.8\right)-4400 \\ =1.3×{10}^{4}N \end{gathered}$$

Now, by using quadratic formula, we get,

$$\begin{gathered} x=\frac{\mathrm{Î\P }Â\pm \sqrt{{\mathrm{Î\P }}^2+2Kk}}{k} \\ =\frac{1.3×{10}^4Â\pm \sqrt{{\left(1.3×{10}^4\right)}^2+\left(2×0.15×{10}^6×4.9×{10}^4\right)}}{0.15×{10}^6} \\ =0.90m \\ =90cm \end{gathered}$$

Hence, the maximum distance x that the spring is compressed is x = 90 cm

Letdistance d' be the distance above the relaxed position of the spring and d'>x . Also, using the bottom-most point as the reference level for computing gravitational potential energy, we get

$$\begin{gathered} \frac{1}{2}k{x}^2=mgd\text{'}+fd\text{'} \\ d\text{'}=\frac{k{x}^2}{2\left(mg+d\right)} \\ =\frac{0.15×{10}^6×0.{90}^2}{2\left(1800×9.8+3.7\right)} \\ =2.8m \end{gathered}$$

Hence, the distance d' that the cab will bounce back the shift is, d' = 2.8 m

Let the elevator come to final rest at the equilibrium position of the spring, with the spring compressed at an amount$d_{eq}$given by

$$\begin{gathered} mg=k{d}_{eq} \\ {d}_{eq}=\frac{mg}{k} \\ =\frac{1800×9.8}{0.15×{10}^6} \\ =0.12m \end{gathered}$$

Computing gravitational potential U = mgy , we get,

$$\begin{gathered} mg\left(d_{eq}+d\right)=\frac{1}{2}k{d_{eq}}^2+f{d}_{total} \\ 1800×9.8×\left(0.12+3.7\right)=\frac{1}{2}×\left(01.5×{10}^6\right)×0.{12}^2+4400×d_{toatl} \\ {d}_{total}=15m \end{gathered}$$

Hence, the approximate total distance that the cab will move before coming to rest is,$d_{total}=15m$.