Question

An elevator and its load have a total mass of $800\text{kg}$. If the elevator, originally moving downward at $10\text{m/s}$, is brought to rest with constant deceleration in a distance of $25\text{m}$, the tension in the supporting cable will be ($g=10{\text{m/s}}^2$)

• A. $8000\text{N}$
• B. $6500\text{N}$
• C. $11200\text{N}$
• D. $9600\text{N}$

Answer 1

The correct option is D $9600\text{N}$

Given
$u=10\text{m/s}$
$s=25\text{m}$
Let deceleration downwards be ${}^{\prime }{a}^{\prime }$
$\Rightarrow v^2-u^2=-2as$
$\Rightarrow 0-100=-2a\times \left(25\right)$
$\Rightarrow a=2{\text{m/s}}^2$
A elevator is moving downward with $a=-2{\text{m/s}}^2$
So $T=m(g-a)$
$=800(10-(-2\left)\right)$
$T=9600\text{N}$