Chapter 4, Problem 86GP

To determine

To find : The tension in the table to stop the elevator.

Answer to Problem 86GP

15795 N

Explanation of Solution

Given:

  $\begin{array}{l}\text{Maximum speed, }u=3.5\text{ m/s}\\ \text{Distance, }d=2.6\text{ m}\\ \text{Mass of elevator, }m=1300\text{ kg}\end{array}$

Formula Used :

Tension is given as,

  $T=mg+ma$

Equation of kinematics is,

  $v^2=u^2+2ad$

Calculation :

Substituting the values,

  $\begin{array}{l}{v}^2=u^2+2ad\\ 0={\left(3.5\right)}^2+2a\times 2.6\\ a=-\frac{12.25}{5.2}\\ \text{Now,}\\ T=mg+ma\\ T=1300\times 9.8+1300\times 2.35\\ T=15795\text{N}\end{array}$