A block of 2 kg is suspended from the ceiling trhough a massless spring of spring constant k=100 N/m. What is the elongation of the spring? If another 1 kg is dded to the block, what would be the further eleongation?

Explanation

Let the acceleration of the 3 kg mass reltive to the elevator is a in the downward direction.
From the free body diagram,
`T-1.5g- 1.5 (g/10)+3a=0`
and T-3g-3(g/10)+3a=0`
`rarr T= 1.5g+1.5g+1.5(g/10)+1.5a`………..i
and `T=3 g+3(g/10)+3a=0`......ii

Now equation i xx2
6g+3(g/10)+3a=2T`
Equation ii xx1
`3g+3(g/10)+3a=T`
Substracting the above two equations, we get,
`T=6a`
Putting T=6a in equation ii. we get,
`6a=3g+(g/10)-3a`
`rarr 9a=(33g)/10` ltbr `rarr 9=((9.8)33)/10 = 32.34`
` rarr a= 32.34/9 = 3.59`
`:. T=6a= 6xx3.59 = 21.55N`.
`:. T_1 = 2T=2xx21.55=43.1N`
`:. Mass= Weight/g`
` = 43.1/9.8 = 4.39 ~~ 4.4 kg.`