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When two block of masses m(1) and m(2) a...

When two block of masses `m_(1) and m_(2)` are connected by a massless string passing over a frictionless pulley and `m_(1)` is held on frictionless table and `m_(2)` is suspended from other end as shown in the figure .

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If it is allowed to move freely, then it moves with acceleration 'a' and tension 'T' acts on the string We have to find accelertation and tension in the string .
For `m_(1)` net force in X - direction (horizontal) is the tension T, FBD of `m_(1)` is

`T= m_(1)a " "` ...(i)
`R = m_(1)g " "` ...(ii)
For `m_(2)` , FBD is
Net force in Y - direction (vertical ) is `m_(2)g - T`
According to Newton's second law,
`m_(2)g - T = m_(2)g " " ` ...(iii)
Put the value of T from (i) in (iii)
Then,
` m_(2)g - m_(1)a = m_(2)g `
`rArr m_(2)g (m_(1) +m_(2))a `
` rArr a = (m_(2))/((m_(1) +m_(2))) g`
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