vf=vi-2at but a= force/mass= 2*8e4/2000=80 m/s^2
so figure time t, if vf=0
distance: avg speed= vi/2, so distance= avgspeed*time
Among the safety features on elevator cages are spring-loaded brake pads which grip the guide rail if the elevator cable should break. Suppose that an elevator cage of 2000 kg has two such pads, arranged to press against opposite sides of the guide rail, each with a force of 8.00 104 N. The friction coefficient for the brake pads sliding on the guide rail is 0.15. Assume that the elevator cage is falling freely with an initial speed of 10.5 m/s when the brake pads come into action. How long will the elevator cage take to stop (in seconds)? How far will it travel (in m)? How much energy is dissipated by friction? (in J)
1 answer
1 answer