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))`.

Options

A. `T_(1) = 50N, T_(2) = 38N`
B. `T_(1) = 35.4N, T_(2) = 94.4N`
C. `T_(1) = 94.4N, T_(2) =35.4N`
D. `T_(1) = 0 N, T_(2) = 35.4 N`

Answer

C

Explanation

Given `m_(1) = 5kg, m_(2) = 3kg`
`g = 9.8m//s^(2)` and `a =2m//s^(2)`
For the upper block `T_(1) -T_(2) -5g =5a`
`rArr T_(1) -T_(2) =5(g +a)`
For the lower block `T_(2) -3g =3a`
`rArr T_(2) =3 (g +a) =3 (9.8 +2) =35.4N`
From Eq (i) `T_(1) =T_(2) + 5(g +a)`
`= 35.4 + 5(9.8 +2) = 94 .4 N`
.