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You are designing an elevator for a hosp...

You are designing an elevator for a hospital. The force exerted 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 accelerates upwards with constant acceleration for a distance of 3.0 m and then starts to slow down. What is the maximum speed (in `ms^(-1)`) of the elevator?

Text Solution

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Consider the forces on the person:
`sum F_(y) = m a_(y)`
N - mg = ma, N = 1.6 mg
So, `a = 0.60g = 6 ms^(-2)`
`v^(2) = u^(2) + 2as = 0^(2) + 2 xx 6 xx 3 implies v = 6 ms^(-1)`
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