Question

The maximum force exerted on a passenger by the floor of the elevator is not to exceed $1.6$ times the weight of the passenger. The elevator accelerates upward with constant acceleration for $3m$. Find the maximum velocity of the elevator.

• A. $4m{s}^{-1}$
• B. $5m{s}^{-1}$
• C. $6m{s}^{-1}$
• D. $7m{s}^{-1}$

Answer 1

The correct option is C $6m{s}^{-1}$

Given : $u=0$ m/s $S=3$ m $g=10$ $m/s^2$

Let the acceleration and the maximum velocity of the elevator be $a$ and $v$, respectively.

Vertical direction : $ma=1.6mg-mg$

$\therefore$ $ma=0.6mg$ $⟹a=0.6g=0.6\times 10=6$ $m/s^2$

Using $v^2-u^2=2aS$ (with initial velocity zero)

$\therefore$ $v^2-0=2\times 6\times 3$ $⟹v=\sqrt{36}=6$ $m/s$