Question #297578

A 2.5 KG MASS AND A 4 KG MASS ARE ATTACHED TO A LIGHT WEIGHT CORD THAT PASSES OVER A FRICTIONLESS PULLEY. THE HANGING MASSES ARE LEFT FREE TO MOVE. FIND THE

A. ACCELERATION

B. TENSION OF THE CORD


1
Expert's answer
2022-02-16T08:34:27-0500

Given:

m1=2.5kgm_1=\rm 2.5\: kg

m2=4kgm_2=\rm 4\: kg


(a) The acceleration of system


a=gm2m1m2+m1=9.84.02.54.0+2.5=2.3m/s2a=g\frac{m_2-m_1}{m_2+m_1}=9.8*\frac{4.0-2.5}{4.0+2.5}=2.3\:\rm m/s^2


(b) The tension in cord

T=m2(ag)=4(9.82.3)=30NT=m_2(a-g)=4(9.8-2.3)=30\:\rm N


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