17. (a) The coin undergoes free fall. Therefore, with respect to ground, its acceleration is G G acoin = g = (−9.8 m/s 2 )ˆj.
(b) Since the customer is being pulled down with an acceleration of G G ′ acustomer = 1.24 g = (−12.15 m/s 2 )ˆj, the acceleration of the coin with respect to the customer is G G G ′ arel = acoin − acustomer = (−9.8 m/s 2 )ˆj − (−12.15 m/s 2 )ˆj = (+2.35 m/s 2 )ˆj.
(c) The time it takes for the coin to reach the ceiling is t=
2h 2(2.20 m) = = 1.37 s. 2.35 m/s 2 arel
(d) Since gravity is the only force acting on the coin, the actual force on the coin is G G G ˆ Fcoin = macoin = mg = (0.567 × 10−3 kg)(−9.8 m/s 2 )ˆj = (−5.56 × 10−3 N)j.
(e) In the customer’s frame, the coin travels upward at a constant acceleration. Therefore, the apparent force on the coin is
G G ˆ Fapp = marel = (0.567 ×10−3 kg)(+2.35 m/s 2 )ˆj = (+1.33 ×10−3 N)j.
