A lift is going up, the total mass of the lift and the passenger is 1500 kg. The variation in the speed of the lift is as shown in (a) What will be the tension in the rope pulling the lift at, (i) 1s (ii) 6 s (iii) 11 s (b) What is the height to which the lift takes the passenger? (c) What will be the average velocity and acceleration during the course of the entire motion? (g=9.8m//s^(2))

A lift is going up, the total mass of the lift and the passenger is 15

1.	A lift is going up, the total mass of the lift and the passenger is 1500 kg. The variation in the speed of the lift is as shown in (a) What will be the tension in the rope pulling the lift at, (i) 1s (ii) 6 s (iii) 11 s (b) What is the height to which the lift takes the passenger? (c) What will be the average velocity and acceleration during the course of the entire motion? (g=9.8m//s^(2))
Answer» Solution :(a) As slope of `v//t` graph gives acceleration, so (i) At `t=1s` `a=(3.6-0)/2=1.8m//s^(2)` i.e, lift is acc. Up, so `T=m(g+a)` i.e, `T=1500(9.8+1.8)` or `T=17400 N` (II) At `t=6` s `a=(3.6-3.6)/(10-2)=0m//s^(2)` i.e., lift has no acc,, so `T=mg` i.e, `T=1500xx9.8` or `T=14700N` (iii) At `t=11` s `a=(0-3.6)/(12-10)=-1.8m//s^(2)` i.e, lift is acc. down, so `T=m(g-a)` i.e, `T=1500(9.8-1.8)` or `T=12000N` (b) The displacement of the passenger is equal to the area under `v//t` curve i.e, `=1/2(2-0)xx3.6+(10-2)xx3.6+1/2(12-10)xx3.6` `=3.6[1+8+1]=36m` `"AVERAGE velocity"=("displacement")/("time TAKEN")` `=36/12=3m//s` `"and Average acceleration"=("CHANGE in velocity")/("time taken")` `=(0-0)/12=0`