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

Options

A. 5.89 N
B. 57.1 N
C. 6.71 N
D. none of these

Answer

B

Explanation

(b) Given, `u=0ms=15 m,t=0.4s`
From `s=ut(1)/(2) at^(2) implies 1.5 =0+(1)/(2)a(0.4)^(2)`
`implies a=(1.5xx2)/((0.4)^(2))=18.75 ms^(-2)`
As the string is moving upwards with in acceleration
`therefore T=m(g+a)=2(9.8+18.75)=57.1 N`