What is the efficiency of the movable unit if, to lift a load weighing 50 kg to a height of 4 m, a force of 300 N

What is the efficiency of the movable unit if, to lift a load weighing 50 kg to a height of 4 m, a force of 300 N was applied to the free end of the rope and it moved 5 m?

Solution:
Efficiency = mgh / FS * 100% = 2000/1500 * * 100% = 133.33%
Answer: 133.33%



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