an elevator starts fro rest with constant upward accl.. it moves 2m in the first 0.6s. a passenger in the elevator is holding a 3kg package by a vertical string. what is the tension in the string during accl. (g=9.8 m/s square

Asked by  | 28th Jul, 2012, 07:32: AM

Expert Answer:

The acceleration of the elevator can be found out by using 2nd equation of motion:
initial velocity of elevator, u = 0m/s
diatance traveled in time t= 0.6 sec is s = 2m
now in the elevator, the total accleleration on the box will be in downward direction and hence it is : a+g = 10.91
thus tension in string = 10.91*3 = 32.73N

Answered by  | 28th Jul, 2012, 02:52: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.