A movable pulley of negligible mass carries a block of mass m. One end of a light inextensible string is fixed to the roof of a lift; the string passes under the movable pulley and its other end is attached to the lower end of a vertical spring of force constant k whose upper end is also fixed to the lift. The lift accelerates vertically upward with acceleration a. The equilibrium extension of the spring is:
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Question

A movable pulley of negligible mass carries a block of mass m. One end of a light inextensible string is fixed to the roof of a lift; the string passes under the movable pulley and its other end is attached to the lower end of a vertical spring of force constant k whose upper end is also fixed to the lift. The lift accelerates vertically upward with acceleration a. The equilibrium extension of the spring is:

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G1 Physics · Accepted Answer

In the upward-accelerating lift frame, effective gravity is g+a downward. At equilibrium the movable pulley and attached mass are supported by two equal string tensions, so 2T = m(g+a). Since the string tension equals the spring force T=kx, we get 2kx=m(g+a), hence x = m(g+a)/(2k).