Workers are loading equipment into a freight elevator at the top floor of a building. However, they overload the elevator and the worn cable snaps. The mass of the loaded elevator at the time of the accident is 1600 kg. As the elevator falls, the guide rails exert a constant retarding force of 3700 N on the elevator. At what speed does the elevator hit the bottom of the shaft 72 m below?

To obtain the weight of the elevator W, substitute 1600 kg for mass of the elevator m and 9.81 m/s² for free fall acceleration g in the equation W = (m) (g),
W = (m) (g)
= (1600 kg) (9.81 m/s²)
=15680 kg,m/s²
= (15680 kg,m/s²) (1 N/1 kg,m/s²)
= 15680 N
The magnitude of the net force F will be,
F = W-R
To obtain the magnitude of the net force F, substitute 15680 N for W and 3700 N for retarding force R in the equation F = W-R,
F = W-R
= (15680 N) – (3700 N)
=11980 N
Rounding off to two significant figures, the magnitude of the net force F will be 12000 N.
To obtain the acceleration a, substitute 12000 N for F and 1600 kg for m in the equation a = F/m, we get,
a = F/m
= 12000 N/1600 kg
= (7.5 N/kg) (1 kg.m/s²/1 N)
= 7.5 m/s₂
To obtain the time t to fall, substitute -72 m for y and -7.5 m/s2 for a in the equation t = √2y/a,

t = √2y/a
= √2(-72 m) /(-7.5 m/s²)
= 4.4 s
To obtain the final speed v at which the elevator hits the bottom of the shaft 72 m below, substitute 7.5 m/s² (only magnitude of a) for a and 4.4 s for t in the equation v = at, we get,
v = at
= (7.5 m/s²) (4.4 s)
= 33 m/s
From the above observation we conclude that, the speed at which the elevator hits the bottom of the shaft 72 m below would be 33 m/s.