A lift starts from rest with a constant upward acceleration It moves 1.5 m in the first 0.4 A person standing in the lift holds a packet of 2 kg by a string Calculate the tension in the string during the motion .

Explanation

Here , ` u = 0 , s = 1.5 m , t = 0.4 s `
From ` s = ut + (1)/(2) at^(2) `
` 1.5 = 0 + (1)/(2) a (0.4)^(2) , `
` a = (1.5 xx 2 )/(0.4 xx 0.4 )= 18 . 75 m // s^(2) `
As the string is moving upwards with this acceleration
` :. T = m (g+a) = 2 (9.8 + 18 . 75) = 57 .1 N ` .