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You are designing an elevator for a hospital . the force on a passenger by the floor of the elevator is not to exceed 1.60 times the passenger’s weight. The elevator accelerates upwards with constant acceleration for a distance of 3.0 m and then starts to slow down. What is the maximum speed $(i n m s^{- 1})$ of the elevator? $(g = 10 m / s^{2})$

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Detailed Solution

$\sum F_{y} = m a_{y}; n - m g = m a$

$n = 1.6 m g$ so $a = 0.60 g = 6 m / s^{2}$

$v^{2} = u^{2} + 2 a s$

$\Rightarrow v^{2} = 0^{2} + 2 \times 6 \times 3 = 6 m s^{- 1}$

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