One end of a spring of force constant `k_1` is attached to the ceiling of an elevator. A block of mass `1.5kg` is attached to the other end. Another spring of force constant `k_2` is attached to the bottom of the mass and to the floor of the elevator as shown in figure. At equilibrium, the deformation in both the spring is equal and is `40cm`. If the elevator moves with constant acceleration upward, the additional deformation in both the spring is `8cm`. Find the elevator's accelerationn ( `g=10ms^-2`).

Answer

`(2)`

Explanation

At equilibrium, `(k_1+k_2)x=mg`
`impliesk_1+k_2=(mg)/(2)` (i)
`(x=40cm, x_1=48cm)`
In case of constant acceleration,
`(k_1+k_2)x_1-mg=ma` (ii)
From Eqs. (i) and (ii), we get `a=2ms^-2`