Work done question?

http://www.ocr.org.uk/Images/65364-question-paper-unit-g481-mechanics.pdf

Q3. B(III) & (IV)
On iii when i am calculating the Wd by the resistive force why am i multiplying it by the 18 m/s when it is going in the opposite direction of the force? Isn't work done just multiplied by the distance the force is acting in?

And on iv why am i adding the resistive force to the 1100n? Is it because the the velocity is constant and there's no acceleration so there must be an equivalent 300N force acting up the slope as well?

1. Yeah W=Fd => W =18*300 = 5400 J. So that's 5400 J/s against gravity.

2. There is no net force on the car since it is at constant velocity so the force from the engine must be equal to the resistive force+the component of the cars weight down the slope.

ty I get number 2, but on number it is travelling 18m/s in opposite direction of resistive force and work done is force multiplied by force X distance moved in direction of force, and in this case is the 18m/s not in the direction of the resistive force?

Force x distance moved in direction of force is the work done by a force.
The question asks for the work done against the force.