A block rests on the floor of a stationary elevator. The elevator then begins to descend with an acceleration of 12 m/s ^2 . Determine the displacement of the block during the first 0.2 s after the start. Take g = 10 m/s ^2 .

Step-by-step solution

The downward acceleration of the elevator is a = 12 m/s ^2 . Since the downward acceleration of the elevator is greater than the acceleration due to gravity ( a > g ), the normal force becomes zero and the block loses contact with the floor of the elevator. The block will undergo free fall under gravity with an acceleration g = 10 m/s ^2 . Using the second equation of motion for the block: s = ut + 1 2 gt^2 Given the initial velocity u = 0 and time t = 0.2 s : s = 0 + 1 2 10 (0.2)^2 s = 5 0.04 = 0.2 m s = 20 cm