Question

Consider an elevator in a hospital. The elevator accelerates upward with constant acceleration for a distance of $100cm$ and then starts to slow down. Also, it is known that the force exerted on a passenger by the floor of the elevator does not exceed $1.8$ times the passenger's weight. Now, find the maximum speed of the elevator in SI units: Take $g=10m{s}^{-2}$ .

Answer 1

Answer: 4

Consider the forces on the person:
$\Sigma F_y=m{a}_y$ $\Rightarrow R-mg=ma$ $\Rightarrow R_{max}=1.8mg$
$\Rightarrow a_{max}=0.8g=8m{s}^{-2}$ , using the third equation of motion,
$v_{max}^2=u^2+2a_{max}$ $\Rightarrow v_{max}^2=0^2+2\times 8\times 1$
$\Rightarrow v_{max}=\sqrt{16}=4m{s}^{-1}$