# What is the difference between work done against gravity and work done by gravity?

Work done "BY" a force,from my understanding,is:

•positive when the direction of displacement is same as the direction of force.

•negative when the direction of displacement is opposite to the direction of force.

Now,work done "AGAINST" any force, intuitively, must be equal to the resultant force,i.e.,(applied force -resistive force)×displacement in the direction of the applied force.

Hence,is the following equation correct?

In pulling a body vertically upwards,

Work done against gravity=Work done by gravity(in opposing the motion of the body)+change in potential/kinetic energy of the body.

"Work done against a force" is not a useful concept. Forces do work; you can find the work any force does over some path.

Work "against a force" just doesn't make sense, since there could be various other forces present that have components "against" the force in question. I have seen two interpretations before. It's either the work done by a force that points in the opposite direction, or it's just the negative of the work done by the force in question. Unfortunately, as you have noted, these are not necessarily equal.

If you see this term, I suggest finding a less ambiguous clarification.

From my understanding, work done against gravity would be work that is applied in opposition to gravity (the opposite direction the gravity is pulling towards); whereas work done by gravity would simply be in the direction gravity is pulling towards.

Lets say you have a ball with a mass hanging by a rope, then the work done by gravity would just be mgh (m being mass, g the gravitational acceleration, and h being the height), from U (potential energy) = mgh. The work done against gravity in order for the ball to stay stationary would have to be an equivalent force in the opposite direction, -mgh. Therefore, in this case the work done against gravity = work done by gravity, which is the case of a simple stationary system.

However, even if you were to drop a ball from a point A to a final position of B, with a displacement of r for instance, (because of gravity being a conservative force, the work done would just be the work done from point A to point B) the work done by gravity on the ball would be the force times the dot product of the change in displacement, as from:

$$\int_A^B F \bullet dr = mgh = mg (A-B)$$

and if this is a closed system where energy is conserved,

$$K_1+U_1=K_2+U_2$$

where $$K_1$$, $$K_2$$ are the initial and final kinetic energies respectively, and $$U_1$$ and $$U_2$$ are the initial and final potential energies respectively, so the equation stated by you above would technically be correct, except that there would be no change in mechanical energy if energy is conserved, work done against gravity = work done by gravity by your definition.