In order to calculate work done against force of gravity we consider we have to apply a force of $F=mg$ on a object and we have to do work $W=mgh$. If gravitational force is also $mg$ how can an object be displaced against gravity with same amount of force? Both forces are equal and they are opposite in direction.
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3 Answers
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In order to begin lifting (vertically) an initially resting object, the lifting force must be greater than mg in magnitude. This allows the velocity to change from zero to upward greater than zero. That means positive work is done.
If the initial lift is only mg, the object will not begin moving upward.
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If you move the object very slowly you need only a very small acceleration and therefore only al title more than mg and this only for a small way, and if you calculate without friction , the object has in the hight h some (very small) kinetic energy. Usually you neglect this energy and the acceleration, so you come up with the energy mgh as a lower bound for the work. (In reality , friction will ad more to the work than the short and small acceleration in the beginning.)
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Well it wouldn't. The equal and opposite force to F=mg is the normal force of the ground, and obviously the object stays still because net force is 0 in the vertical axis. If you want to raise the object, you then apply an additonal upward force, and the equal and opposite force is felt as the object pushing back. But because you are more massive, or standing on the ground, that force barely displaces you downward, while the upward force you applied in additon to the normal force has greater displacemet for the less massive object in question.
