Question

You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is not to exceed 1.70 times the passenger's weight. The elevator accelerates upward with constant acceleration for a distance of 2.3 m and then starts to slow down. What is the maximum speed of the elevator?

Answer 1

Answer:

5.62 m/s

Explanation:

Newton's law of motion can be used to determine the maximum speed of the elevator. In the question, we are given:

Force exerted by the elevator (R) = 1.7 times the weight of the passenger (m*g)

Thus: R = 1.7*m*g

Distance (s) = 2.3 m

Newton's second law of motion: R - m*g = m*a

1.7*m*g - m*g = m*a

a = 0.7*m*g/m = 0.7*g = 0.7*9.8 = 6.86 m/s²

To determine the maximum speed:

$v^{2} _{f} = v^{2} _{i} + 2as$$= 0 + 2(6.86)(2.3) = 31.556$

$v_{f} = \sqrt{31.556} = 5.62 m/s$

Therefore, the elevator maximum speed is equivalent to 5.62 m/s.