How does $W=0$ for the following equation and question on conservative forces? We have $W = fd\cos(\theta)$ = Kinetic Energy + Potential Energy.
Our example is a $10$kg ball is falling from a height of $10$m. I can see why Kinetic Energy and Potential Energy cancel out to become $0$, but how does $
W=fd\cos(\theta)$ come to 0 when theta is equal to $180$ degrees as it is falling, so $\cos(180)$ is $-1$ and $d=10$, and $f= 10g$. This does not equal 0. Thanks for the help I am probably making a mistake.
 A: If you consider the ball alone as the system then the system (the ball) only has one force acting on it which is the force due to the gravitational attraction of the Earth $ \vec f = m \vec g$ where $m$ is the mass of the ball and $ \vec g$ is the gravitational field strength which is in a downward direction.
The assumption is that there is no air resistance.
If the ball is falling down and the displacement of the ball is $\vec d$ in the downward direction then the work done by the external force is $\vec f \cdot \vec d = fd \cos(0) =fd$ as both the force and the displacement are in the same direction.
In the process of falling the change in kinetic energy of the ball is $\Delta E_{\rm k} = fd$, the work done on the ball.  
Now consider the ball and the Earth as the system with no external forces acting on the system and there being no air resistance.
Unlike the ball alone system, this system possesses gravitational potential energy which depends on the masses of the Earth and the ball and their separation.
As the ball is moving down towards the Earth (and the Earth is moving “upwards” towards the ball) the Earth and the bake gain kinetic energy and the system (ball and Earth) loses an equal amount of gravitational potential energy with the change in total energy of the system being zero.
Because the mass of the Earth is so much greater the the mass of the ball the change in kinetic energy of the Earth is often neglected as it is so much smaller than the change in kinetic energy of the ball.
A: You have made no mistake other than forgetting that the work done by gravity is already included. It is just given the name "potential energy".
A: The net work done in a gravitational field is zero for a closed loop. For example, the net work done would be zero if you considered the work by an external force to raise the mass 10 m from the ground (which increases its potential energy) plus the work done by gravity when the mass falls from 10 m to the ground (which decreases its potential energy in exchange for kinetic energy) making the sum zero. The work you calculated only covers the latter. The negative sign simply means that the work done by gravity decreases potential energy. 
By the way, the force of gravity is a conservative force, which means the work done in moving between two points is independent of the path taken between the two points.
A: The "mistake" is that
$\Delta E_k + \Delta E_p  \neq W_{Total}$
It is 
$\Delta E_k + \Delta E_p  \neq W_{non\ conservative}$
Since your forces are conservative, you find that $W_{NC}=0$, but that doesn't mean there's no total work. The conservative woork is precisely $E_p$. Check another answer of mine for clarification:
https://physics.stackexchange.com/a/420447/157625
