\(\text{Consider a block placed on the floor of a lift moving upwards with a}\)

\(\text{constant velocity.}\)

\(\text{In frame of lift the block appears stationary and the earth appears to }\) \(\text{move down, Is the mechanical Energy of block and earth system }\) \(\text{conserved in lift frame of reference..??}\)

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## Comments

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TopNewest@rohit Gupta am i correct

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First of all, an independent frame of reference is in free motion, i.e., either in uniform motion unaffected by forces, or it is falling. The life "moving upwards with a constant velocity" is neither, since it's being hauled up and work is being done. In this case, use the frame of reference that's based on the center of gravity of both the Earth and the lift with the block. Then we see that work is being done to separate the two, much like pulling apart two items connected by a spring.

It is a convenient approximation to assume that Earth's mass is infinite, so that its center of gravity is the same as the center of gravity of the system. Then we can analyze the behavior of smaller items moving around in Earth's gravitational force field, using Earth's frame of reference. Trying to do this the other way around is at best problematic. It isn't symmetrical.

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energy can not be conserved. here ke of block + earth system is constant. Work done by graviational by lift on earth and block system, as well as work done by gravitational force on earth by block will be equal to work done by tension force exerted by lift system( string will exert force on ceiling by which lift is suspended... . and ceiling is part of earth)

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The tension wont do any work in frame of lift... As in frame of lift, lift remains at rest and hence no displacement..!!

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tension will do some work!! In frame of lift, it can be easily visualized that tension is the force which is bringing ceiling or roof nearer to the lift!! i think u can visualize the situation

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