91Ó°ÊÓ

• Mass of the passengers = 1700 kg.
• Acceleration = 1.20 m/s²

Apply Newton’s Second Law of motion as:

${\mathit{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathit{m}\mathit{a}$

$\mathit{T}\mathbf{-}\mathit{m}\mathit{g}\mathbf{=}\mathit{m}\mathit{a}$ …â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(¾±)

Here, m is the mass of the passenger, T is the tension in the cable, g is the acceleration due to gravity, and F_net is the net force exerted.

Substitute 1700 kg for m, 9.8 m/s² for g, and 1.20 m/s² for a in the above equation, and we get,

$\begin{array}{c}T−1700\text{ k²µ}×9.8{\text{ m/²õ}}^{\text{2}}=1700\text{ k²µ}×1.20{\text{ m/²õ}}^{\text{2}}\\ T−16660\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}=2040\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}\\ T=\left(2040+16660\right)\text{ N}\\ T=18700\text{ N}\end{array}$

Hence, the tension in the cable is 18700 N.

Since the elevator moves with constant velocity, the acceleration is zero.

Substitute 1700 kg for m, 9.8 m/s² for g, and 0 for a in equation (i), and we get,

$\begin{array}{c}T−1700\text{ k²µ}×{\text{9.8m/s}}^2=1700\text{kg}×0\\ T=16660\text{ k²µ}â‹\dots {\text{m/s}}^2\\ T=16660\text{ N}\end{array}$

Hence, the tension in the cable is 16660 N.

Substitute 1700 kg for m, 9.8 m/s² for g, and (-0.6) m/s² for a in equation (i), and we get,

$\begin{array}{c}T−1700\text{ k²µ}×{\text{9.8m/s}}^2=1700\text{ k²µ}×\left(−0.6\right){\text{ m/²õ}}^2\\ T−16660\text{ k²µ}â‹\dots {\text{m/s}}^2=−1020\text{ k²µ}â‹\dots {\text{m/s}}^2\\ T=\left(16660−1020\right)\text{ N}\\ T=15640\text{ N}\end{array}$

Hence, the tension in the cable is15640 N.

Calculate the distance covered in the first step as:

$y_1=u{t}_1+\frac{1}{2}{a}_1{t}_1^2$

Here $y_1$ is the distance covered in the first step, u is the initial velocity, and t₁ is the time.

Substitute1.20 m/s² for a₁, 0 for u, and 1.50 s fort₁ in the above expression, and we get,

$\begin{array}{c}{y}_1=\frac{1}{2}×1.20{\text{ m/²õ}}^2×{\left(1.5\text{ s}\right)}^2\\ =1.35\text{ m}\end{array}$

Calculate the velocity after the first step can be written as:

$v_1=a_1{t}_1$

Here, v₁ is the final velocity after the first step.

Substitute1.20 m/s² for a₁ and 1.50 s fort₁ in the above expression, and we get,

$\begin{array}{c}{v}_1=1.20{\text{ m/²õ}}^2×1.50\text{ s}\\ \text{=1.80 m/²õ}\end{array}$

Calculate the distance covered in the second step as:

$y_2=v_1{t}_2+\frac{1}{2}{a}_2{t}_2^2$

Substitute1.80 m/s for v₁, 0 for a₂, and 8.50 s for t₁ in the above expression, and we get,

$\begin{array}{c}{y}_2=1.80\text{ m/²õ}×\text{8.50 s}+0\\ =15.3\text{ m}\end{array}$

Calculate the distance covered in the third step.

$y_3=v_2{t}_3+\frac{1}{2}{a}_3{t}_3^2$

Substitute1.80 m/s for v₂, (-0.6) m/s² for a₃, and 3.0 s for t₃in the above expression, and we get,

$\begin{array}{c}{y}_3=1.80\text{ m/²õ}×3.0\text{ s}−\frac{1}{2}×0.6{\text{​m/²õ}}^2×{\left(3\text{ s}\right)}^2\\ =5.4\text{ m}−2.7\text{ m}\\ =\text{2.7 m}\end{array}$

Calculate the final velocity after the third step as:

$v_3=v_2+a_3{t}_3$

Substitute1.80 m/s for v₂, (-0.6) m/s² for a₃, and 3.0 s for t₃ in the above expression, and we get,

$\begin{array}{c}{v}_3=1.80\text{ m/²õ}−0.6{\text{ m/²õ}}^2×3.0\text{ s}\\ =1.80\text{ m/²õ}−1.80\text{ m/²õ}\\ =\text{0 m/²õ}\end{array}$

Calculate the total distance traveled as:

$y_{total}=y_1+y_2+y_3$

Substitute 1.35 m for y₁, 15.3 m fory₁, and 8.10 m for y₃ in the above expression, and we get,

$\begin{array}{c}{y}_{total}=\left(1.35+15.3+2.7\right)\text{ m}\\ =\text{19.35 m}\end{array}$

Hence, the total distance traveled is 19.35 m, and the final speed is 0 m/s.