KaitlynLengyel

__ **Week 07 ( 05th of March, 2012)** __

01 . An object is hung from a spring balance attached to the celing of an elevator cab. The balance reads 65 N, when the cab is standing still. What is the reading when the cab is moving upward, (a) with a constant speed of 8 m/s and (b) with a speed 8 m/s while decelerating at a rate of 2.4 m/s²

a) Tsin(90) + mgsin(270)=ma  a=0m/s2 sin(90)=1 sin(270)=-1 T-mg=0 T=mg The reading stays the same because there is no change in acceleration.

b) Tsin(90) + mgsin(270)=m(-a) T-mg=-ma T=-ma+mg T=-(mg/g)a+mg