physics

An elavater m=800kg has a maximum load of 8 people of 600kg. The elevatorgoes up to 10 stories = 30m at a costant speed of 4m/s. what is the average power output of the elevater motor if the elevater is fully loaded with its maximum weight? (neglect friction)

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asked by Noemi
  1. power= work/time= (600+800)g*30/(time)

    where time is 30/4 seconds

  2. P = F v = w v = (800+600)(4) = 5600 W

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    posted by CANTIUS

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