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In a system with 2 pulleys, to lift a x kg object y meters I would need half the force that the object exerts due to gravity but pull the rope double the distance, why does this happen this way?

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  • $\begingroup$ Think of it like a lever, one part moves more than the other... $\endgroup$ – user207455 Jul 27 at 8:01
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This can be viewed as a requirement of the work-energy theorem and the conservation of energy. Assuming you alter the system slowly enough that parts don't gain a significant kinetic energy, the work you do by pulling the rope must be equal to the work done against gravity or any other forces that may be applicable. If you pull the rope with a force $F$ by a distance $a$, and the object of interest with weight $W$ is lifted a distance of $b$, we have $$ F a \ge W b, $$ where equality holds in idealized systems where work is only being done by pushing the weight $W$ against gravity. For a pulley system, this means having no friction, massless strings and massless pulleys. If $ a = 2b $ in such an idealized system, then $ F = W/2 $.

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