A body of mass $$4$$ kg is suspended by a spring balance from the roof of an elevator cage. When the cage is moving up with constant acceleration, reading of the balance is $$48$$ N. When the cage is moving down with same acceleration, reading of the spring balance is $$g=10$$ $$m/s2$$.

Question

A body of mass $$4$$ kg is suspended by a spring balance from the roof of an elevator cage. When the cage is moving up with constant acceleration,  reading of the balance is $$48$$ N. When the cage is moving down with same acceleration, reading of the spring balance is $$g=10$$  $$m/s2$$.

Infinity Learn · Accepted Answer

$$(3)mg+ma=R1$$⇒   $$4$$  ×  $$10+4$$ ×  $$a=48$$    ⇒  $$a=2$$ $$m/s2Again$$    mg−$$ma=R2$$⇒   $$R2=4$$  ×  $$10$$−$$4$$  ×  $$2$$ $$N=32$$  N.

A body of mass 4 kg is suspended by a spring balance from the roof of an elevator cage. When the cage is moving up with constant acceleration, reading of the balance is 48 N. When the cage is moving down with same acceleration, reading of the spring balance is $(g = 10 m / s^{2}) .$

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A body of mass 4 kg is suspended by a spring balance from the roof of an elevator cage. When the cage is moving up with constant acceleration, reading of the balance is 48 N. When the cage is moving down with same acceleration, reading of the spring balance is g=10 m/s2.

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A body of mass 4 kg is suspended by a spring balance from the roof of an elevator cage. When the cage is moving up with constant acceleration, reading of the balance is 48 N. When the cage is moving down with same acceleration, reading of the spring balance is $(g = 10 m / s^{2}) .$