Here is a great diagram of an inclined plane with all the forces shown,
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http://dunningrb.files.wordpress.com/2007/10/300px-free_bodysvg.png
Print or draw the diagram, so you can draw on it!!
The force needed to pull the 12 kg mass up the inclined plane is labeled f. The weight of the object is mg. It is drawn straight down because gravity pulls straight down. Draw a line from the arrow head at the end of mg to the arrow head at the end of mg sin θ. The angle with vertex at the point where your line and mg intersect is θ. Now you have a right triangle with mg as the hypotenuse, and mg sin θ as the side opposite θ. The triangle that you made and the inclined plane triangle are similar triangles. Later when do problems with friction, you will use mg cos θ.
m = 12 kg
g = 9.8 m/s^2
mg = 117.6 Newton
mg sin θ = 58.8 Newton
mg sin θ pulls the 12 kg block down the plane.
The force f is the force needed to pull the block up the 80 m ramp without accelerating it, so, f = mg sin θ.
f = 58.8 N
distance = 80m
Work = force * distance
Work = 58.8 N * 80 m = 470.4 N-m
At the same time that the block was pulled up the ramp, it was lifted 40m
Change in Potential Energy = mass * gravity * change in height.
ΔP.E. = 12 kg * 9.8 m.s^2 * 40 m
ΔP.E. = 470.4 Joules
Interestingly, the work done to the block = the increase in P.E.