To stop an elevator in free fall, brakes are installed between the walls of the elevator car and the walls of the shaft in which the car moves. In addition, springs with compression coils are installed on the floor, at the bottom of the shaft, in order to absorb the residual energy if the brakes fail to completely stop the elevator car.

Question

There is an elevator car, with 3 people on board, whose cables become unhinged. The elevator falls 15 m. The total mass of the car and the people on board is 4000 kg. The sum of the frictional forces applied by the brakes is 5000 N acting over a distance of 15 m. Each spring attached to the floor has a spring constant of 5.0 x 10^6 N/m and may be compressed over a maximum distance of 30 cm.

How many springs are needed on the floor so that the elevator car and its occupants do not compress them by more than 20 cm?

Answer 1

Geepers: a better question for followon is what deacceleration will happen in that 20 cm stopping distance.

OriginalPEnergy-frictionalenergy=FinalPEsprings.

4000*9.8*15-5000*15=n*5E6*1/2*.2^2

solve for n. I get about six.

Answer 2

Now for the better question: what is the deacceleration?

KEfinal=keinitial-force*distance
4000*9.8*15-5000*15 -5000a*.2

a = about 50 g. Wouldn't that crush bones and flesh? check my thinking.