The cable of the 1800 kgelevator cabin Figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 3.7 m above a spring of spring constant k = 0.15 MN/m . A safety device clamps the cab against guide rails so that a constant frictional force of 4.4 kNopposes the cab’s motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance xthat the spring is compressed (the frictional force still acts during this compression). (c) Find the distance that the cab will bounce back up the shaft. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)

The mass of the cab is, m = 1800 kg

Distance between cab and spring is d = 3.7 m

The spring constant is,

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

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,$x=\frac{\zeta \pm \sqrt{{\zeta }^2+2kK}}{k}$

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,

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

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

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

And

$$\begin{gathered} \zeta =mg-f \\ =\left(1800\times 9.8\right)-4400 \\ =1.3\times {10}^{4}N \end{gathered}$$

Now, by using quadratic formula, we get,

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

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

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

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

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

Computing gravitational potential U = mgy , we get,

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

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