Question

Consider an elevator in a hospital. The elevator accelerates upward with constant acceleration for a distance of $100\,cm$ 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 $g=10\,ms^{- 2}$ .

Answer 1

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