question from the STAV 2013 trial exam

[Chazef]

This involves the STAV trial exam which may be used as a trial exam at your school so I guess this thread contains spoilers?

Anyway for anybody has done it, the question at the beginning of the motion section where it asked for the work done by the car engine, do you do this: (Work done by engine) =  (driving force of the car)*(distance travelled) + (Friction force)*(distance travelled). My friend who is quite the boss said it's supposed to be a '-' not a '+'. I ended up with 480000 joules. Anybody able to shed light on this?

[lzxnl]

OK...let's look at this overall.
Work done ON the car is net force times distance. Net force is (driving force - friction force). So work done ON the car is (driving force - friction force)*distance.
But the work done ON the car is the work done by the engine, so your friend seems right. Even though I don't have the exam with me.

Negative work is possible. Technically, work is a dot product between the force and displacement vectors for a constant force. So if the force vector is directly antiparallel to the displacement, the work done is negative. Makes sense; if you're moving up, and there is a force pushing down on you, your kinetic energy is going to drop. Work just means change in kinetic energy; positive work means more kinetic energy, negative work means lower kinetic energy.

[~T]

I haven't seen the question so I may be misinterpreting this... however, because it is "work done by the car engine" I thought you would discount the frictional force. Wouldn't it be something like:
Net work = Work done by engine - Work done by friction
              = (Driving force)*(Distance travelled) - (Frictional force)*(Distance travelled)

But because we only want the work done by the engine, it would just be (Driving force)*(Distance travelled)?

But the work done ON the car is the work done by the engine, so your friend seems right. Even though I don't have the exam with me.

Doesn't friction do work ON the car?

[lzxnl]

Wait what the...I read the question again and I must have been half-asleep on that night.

Of course. Work done by car engine is just the force the car engine exerts times the distance. So yeah, driving force by distance.

But because we only want the work done by the engine, it would just be (Driving force)*(Distance travelled)?
Doesn't friction do work ON the car?

You're right. I got the concepts confused in my mind and I don't know what happened.