Let's break the question into two parts:
1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.
1. Ramp Scenario:
In an incline, the only component of cart's weight(mg) that is in the direction of motion is [tex]mgsin \alpha [/tex]. Therefore the effort force in this case must be equal or greater than [tex]mgsin \alpha [/tex].
Now we need to find [tex] \alpha [/tex]. [tex] \alpha [/tex] is the angle between the incline of the ramp and the ground.
Since the height is 5m and the length of the ramp is 8m, [tex]sin \alpha [/tex] would be 5/8 or 0.625. Now that you have [tex]sin \alpha [/tex], mutiple it with mg.
=> m*g*[tex]sin \alpha [/tex] = 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is 125N.
2. Lever Scenario:
Just apply "moment action" in this case, which is:
[tex]F_{e} d_{e} = F_{r} d_{r}[/tex]
[tex]F_{e} [/tex] = ?
[tex]F_{r} [/tex] = mg = 20 * 10 = 200N
[tex]d_{e} [/tex] = 10m
[tex]d_{r} [/tex] = 1m
Plug-in the values in the above equation:
[tex]F_{e} [/tex] = 200/10= 20N
As 20N << 125N, the best choice is to use lever.
