Question #269785

click for figure1 https://www.hizliresim.com/94cfjfb

A 1.5-kg

kg block rests on top of a 7.5-kg

kg block. The cord and pulley have negligible mass, and there is no significant friction anywhere.

a)What force F

F must be applied to the bottom block so the top block accelerates to the right at 2.2 m/s

2

m/s2 ?

b)What is the tension in the connecting cord?


1
Expert's answer
2021-11-24T12:42:19-0500

a) The bottom block is the combination of the two masses ;

M=7.5+1.5=9kgM=7.5+1.5=9kg


Therefore the force applied is F=maF=ma

F=9×2.2=19.8NF=9×2.2=19.8N


b) The tension TT in the string is ;


T=1.5×2.2=3.3NT=1.5×2.2=3.3N


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