Sat July 28, 2012 By:

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

Expert Reply
Sat July 28, 2012
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
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