Inclined planes - worked answer please...? Super n00b.

[Cinnah]

Last year I chose physics. I didn't get it because of a subject clash.This year I had to pick up physics BECAUSE of a subject clash.
Hence, I missed out on units 1&2 because the timetabling system hates me.
I am trying my very best to move forward and do the best I can, but understanding all this terminology in such a short time is INSANE. I love the theory, it's amazing, but the application is so grrrrrrrruuuuuuuuuuh.
These holidays have pretty much been me self-teaching all of the simple stuff, with a little help. The stuff my teacher explained on the transition class was useless, because I couldn't take notes on stuff I had no idea about.

Tldr; I am desparate for help on this question.
Nobody has been able to help me so far, and because I don't know how much is new content for this year, I am so afraid of falling behind.
If it helps, I think it's from the Jacaranda textbook (dunnolol, it's a photocopied holiday homework booklet)
the exercise is 1.3 - The normal force and inclined plains

The Question is:
Kirsty is riding in a bobsled that is sliding down an snow covered hill with a slope at 30 degrees to the horizontal. The total mass of the sled and Kirsty is 100 kg. Initially, the brakes are on and the sled moves down the hill at a constant velocity.

3. Calculate the net frictional force on the bobsled.

Now, I've had trouble with this question. Mainly because I remember reading that the frictional force equals the normal force, which is also equal to the mass of the object.
However, I've also read that net force = weight of mass - component force of gravity - force of friction.
...is net force the same as net frictional force?
I know that the acceleration on a plane is g*sine(theta), however that's not useful to this question
also that gravity is 9.8ms^-1, or 10ms^-1, depending on the textbook
I have somehow previously figured out that the bobsled is implying a force of 980N, but I can't remember how.

*takes deep breath* D: Long post is long.
*sits in the dunce corner*
Please. Anybody. Help.

[adelaide.emily10]

i did this question the other day, its from heinemann q.3 of 1.3
-=minus
m=mass
a=accelaration
(-) = theta, the angle
x=multiply

a is 4.9m/s/s because
g x sin (-) = a
9.8 x sin30 = 4.9 m/s/s

Net force = applied force - friction force
net force is zero because it is at a constant velocity

0=(ma) - frict
0= 100a - frict
0= 100 x 4.9 - frict
0=490 - frict
-490 = frict
positive is down therefore, 490 N up the slop = frictional force

[Cthulhu]

Meet my friend the diagram. Please draw them every time you are faced with a force question

[schnappy]

You just need an understanding of forces.

If velocity is constant, then the net force acting on the object is 0, as per newton's laws.  Fnet = ma, v is constant so a=0, Fnet=0.

What does Fnet physically mean? It means any forces acting on an object are cancelled out by other forces. Think if you push an object on one side of it, then if you push it just as strongly (the same force), then you're pushing the object into itself... it won't move because the two pushes cancel each other out. Fnet=0.

So with your question, v is constant so the frictional force that's slowing the object down is the same as the force that is driving the object down your inclined plane. They're two pushes cancelling each other out.

So the frictional force is equal to the diaganol component of weight... something like mg.cos(theta) or something (This should be clear in your textbook somewhere)

Force diagrams should make this explanation a bit clearer.

[Cinnah]

@adelaide.emily10
Thanks so much, this HAS helped.
It's difficult to extract all the formulae for me - I have a running page, but I couldn't seem to find anything to fit the problem

@Cthulhu Fhtagn
A diagram came with the question, and I drew another in my workbook and re-labelled - to no avail.

@shnappy
I've understood the Newtonian laws pretty well, it's just the net force thing that isn't explained in anything I have (or if it is I am very bad at looking), so thanks a bunch.