One end of a spring of force constant k1 is attached to the ceiling of an elevator. A block of mass 1.5 kg is attached to the other end. Another spring of force constant k2 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 springs is equal and is 40 cm. If the elevator moves with constant acceleration upward, the additional deformation in both the springs is 8 cm. The elevator's acceleration (in ms-2) is (g = 10 ms-2).
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Detailed Solution
At equilibrium, k1+k2x=mg⇒ k1+k2=mgx ……..(i) x=40 cm,x1=48 cmIn case of constant acceleration, k1+k2x1−mg=ma ………(ii)From Eqs. (i) and (ii), we get a=2 ms−2