A lift is going up. The total mass of the lift and the passengers is 1500 kg . The variation in the speed of the lift is given by the graph shown in Fig . What will be the tension in the rope pulling the lift at (i) `t = 1 s` , (ii) `t =6 s` (iii) `t = 11 s` ?
What is the height to which the lift takes the passengers ? During the course of entirs motion What is the average velocity and average acceleration of the lift ? Taken `g = 9.8 m//s^(2) `
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Explanation

Here , ` m = 1500 kg , T = R = ? `
(i) From 0 to 2 s , u = 0, `upsilon = 3.6 m //s`,
`a = (upsilon-u)/t =(3.6 - 0 )/0 = 1.8 m// s^(2) `
As acceleration is unifrom therefore at t = 1 s ,
` a = 1.8 m//s^(2) `
`:. R= m (g+a) = 1500 (9.8 + 1.8) `
`1500 xx 11.6 = 17400 N`
(ii) From the graph we find that at t = 6 s,a=0 .
`:. R = mg = 1500 xx 9.8 N = 14700 N `
(iii) At `t = 11 s` :
From the graph the velocity of body is decreasing at a constant rate , From `t = 10 s` to `t = 12 s`
`a = (0-3.6)/(2) = - 1.8 m//s^(2)`
This is the retardation at `t = 11 s `
`:. R = m (g+a) = 1500 (9.8 - 1.8)`
`12000 N`
The height to which the lift takes the passengers = area of trapezium oABC
`=(( AB + OC )/(2))AD = (8+2)/2 xx 3.6 = 36 m`
` =(( AB + OC )/(2))AD = (8+2)/2 xx 3.6 = 36 m`
As initial velocity of lift =0 and final velocity of lift is also zero , thereforce , change in velocity of lift , Delta upsilon = 0 . Average acceleration of lift ,
`a_(av) = (Delta upsilon)/(t) = (0)/(2)` = Zero
Further in 12 secound, net displacement is zero . Therefore , average velocity is zero.