# Confusion about the total work done on an object applied by an upward force

Suppose we have an object which the mass is $$m$$ kg. If we lift up the object and the object moves with constant upwards velocity, the force we're applying on it will be equal to $$mg$$, and say if the object moved a distance of $$h$$ meters, technically we should have done $$mgh$$ joules of work. However, gravity is also doing work on the object, and since its direction opposes to the direction of the object's movement, it should have done $$-mgh$$ joules of the work, and by calculation the total work done on the object should have been $$mgh + (-mgh) = 0$$ joules.

Nevertheless, if we consider the potential energy of the object, since the object has gone up by $$h$$ meters, the potential energy should increase by $$mgh$$ joules as the kinetic energy is kept constant, meaning that there has to be some total work done on the object. So why did we run into a contradiction?

The issue is that if you are taking into account the change in gravitational potential energy, then you have also already taken into account the work done by gravity.

For any conservative force, the work done by that force over some path $$C$$ is equal to the negative change in potential energy from the starting position of the path $$\mathbf a$$ to the ending position of the path $$\mathbf b$$. This is due to the definition of potential energy $$U$$ of a conservative force $$\mathbf F$$

$$\mathbf F=-\nabla U$$

and use of the definition of work with the fundamental theorem of calculus

$$W=\int_C\mathbf F\cdot\text d\mathbf x=\int_C(-\nabla U)\cdot\text d\mathbf x=-(U(\mathbf b)-U(\mathbf a))=-\Delta U$$

Therefore, you can either forget that gravity is conservative and just calculate the total work done on the object (work done by you and gravity), or you can take potential energy into account and then just consider the work done by you. Either way you get the same thing happening. No contradiction.

More specifically, keeping track of energy, if you consider the change in potential energy we have $$\Delta E=\Delta K+\Delta U=W_\text{ext}$$ Since $$\Delta K=0$$, $$\Delta U=mgh$$, and $$W_\text{ext}=W_\text{lift}$$, we have $$mgh=W_\text{lift}$$

or $$W_\text{lift}-mgh=W_\text{lift}+W_\text{grav}=W_\text{net}=0$$

It all works out.

... the potential energy should increase by $$mgh$$ joules as the kinetic energy is kept constant, meaning that there has to be some total work done on the object.

To specifically address this point, there is nothing that says a change in potential energy means the net work done on an object is not $$0$$. This example is clearly a counterpoint to this claim.