Two masses of `5kg` and `3kg` are suspended with help of massless inextensible strings as shown in figure. Calculate `T_(1)`and `T_(2)` when whole system is going upwards with acceleration `=2m//s^(2) (use g = 9.8 ms^(-2))`.

Explanation

Here, `m_(1) = 5kg`
`m_(2) = 3kg`
` g = 9.8 m//s^(2)`
`a = 2 m//s^(2), ` upwards
`T_(1) = (m_(1) +m_(2)) (g +a) = (5+3) (9.8+2) = 94.4 N`
` T_(2) =m_(2) (g +a) = 3 (9.8 +2 ) = 35.4 N ` .