Another Physics question

A car of mass 800kg has a max speed of 20ms-1 down a slope inclined at 2 degrees to the horizontal. The frictional resistance to motion of the car is constant, and of magnitude 1000N. take g =9.81ms-2

a)find power of cars engine (answer: 14400W)
b)find its maximum speed up the same hill (answer: 11.3ms-1)

How do i do these questions, so stuck!

Do you know a (very simple) formula for the power of the engine in terms of the speed of the car and the force provided by the engine.
It should be in your notes.
If so, you first have to find the force provided by the engine. (You know the speed of the car)
The force provided by the engine is the frictional force (given) less the component of the car's weight acting down the slope (which assists the car).
Do you know how to find the component of the car's weight, mg, down the 2 degree slope?


PS
Please give your posts a more meaningful title. "Physics problem" doesn't help as all the posts here are physics problems. You will get more chance of a reply if you do this.

So P=FV

=20000

and then 10000-800gsin(2) = 9726

is that the correct calculation?

if so what next?

Not quite correct.
Friction is 1,000N not 10,000N in the question.

When you've done the subtraction

Then use Power = Fv (with this force) to find power.

By the way. The answer given looks like it's used 10m/s/s for g, not 9.81

you're an absolute legend, so i knew how to do it, but the worksheet had mistakes on it! How would i get max speed?

.
Now you know the power of the car you can apply the same formula P=Fv for the uphill motion to find v. You need to think it through a little for yourself now. What is the force?

So P of engine = 14400
F= 720.8+8000sin2

and then *72%

=10.3? but that's incorrect

The force of the engine in the first part was not calculated like that.
Then it provided the frictional force less the component of the weight down the slope.
Now what does it do?

I'm completely lost

Apply the same formula
P=Fv to the motion up the slope.
You now know the value of P from part 1.
You need to find the value of F (the force from the engine) to plug in to this to get a value for v.
The force the engine required in the first part was the frictional force less the component of the weight down the slope because that was assisting the motion.

This time the car is going up the slope so the component of the the weight is acting against the motion. So the car now needs to provide a force greater than the frictional force.

Why did you add the component of the weight to 720N? This is not the frictional force.

so..

F+8000sin2=720.8?

What does the question say the frictional force is?

1000N

So how much MORE force does the car engine need to move the car up the slope against this and against the component of its weight down the slope (which you already know)?

Use this force in P=Fv to find v

I think you've had enough help on this one.