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Find the work done by tension (in J) on 1 kg block when 4 kg block moves down by 20 cm after the system is released. All surfaces are frictionless

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The correct Answer is:
5
`-2T_S_(4) +TS_(1)=0`
`S_(1)=2S_(4)=0.4 m`
`40-2T=4a_(4)`
`T-mg sin 30=1a_(1)`
`a_(4)=10-T/2`
`a_(1)=2a_(4)`
`T-5 20 -T`
2T=25
`omega_(T)=Txx0.4=12.5/5=5J`
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