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A force of 0.5 N is applied on upper blo...

A force of `0.5 N` is applied on upper block as shown in figure. The work done ( in joule ) by upper block on lower block for a displacement of `3 m` is `:`

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Force of friction between the two blocks, Figure, `f= mu R= mu m g =0.1 xx 1 xx10 = 1N`
As applied force `F=0.5 N ltf.`
`:.` The two blocks move together.
Common acceleration,
`a=(F)/(1+2)=(0.5)/(3)=(1)/(6)m//s^(2)`
`:.` Force applied by upper block on lower block
`F^(') = ma= 2 a = 2xx(1)/(6)=(1)/(3)N`
Work done by upper block on lower block for displacement of `3m = F 3 ^^(') xxs=(1)/(3)xx3= 1J`
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