mushroom25..it's about the same length of time since I did any of this but I think there are a couple of errors in your calculation.
Firstly the weight component down the slope is mgsin(angle of slope)
For a 6% slope the sine of the angle of the slope is approximately 0.06, so in rounded numbers the weight component down the slope is in the order of 46000kg x 10 x 0.06 =27000kg which when you factor in the coefficient of friction of 0.6 becomes 46000kg!
Secondly work = force x distance moved. Lets say you want to move the truck 20m then the work is 46000kg x 20m = 0.9MJoules.
A Joule is a watt per second so if you wanted to take 60s to move the truck 20m up the slope you would have to supply 0.92MJoules in 20s or 920,000/20 = 46kW i.e. about a 50HP motor.