Highest velocity reached on a hill

[Drizzy]

if the top is 60 then the bottom is 30 degrees!

 

[Drizzy]

what do I do next?

 

[PeroK]

what do I do next?

You need to start doing more of the work on this! I thought your idea was to calculate the acceleration down the slope?

 

[Drizzy]

okay so the force down is F1 and the "resulting" force is:

F1-force of friction= m*a

F1= sin(30)*mg

I get that part but then my teacher wrote (2) that the frictional force is equal to m*a2

w8... i think I am getting it.. so on the horizontal fround there won't be a force forward so the only force is the frictional force`?

 

[PeroK]

okay so the force down is F1 and the "resulting" force is:

F1-force of friction= m*a

F1= sin(30)*mg

I get that part but then my teacher wrote (2) that the frictional force is equal to m*a2

w8... i think I am getting it.. so on the horizontal fround there won't be a force forward so the only force is the frictional force`?

You seem to have a habit of writing one line and then stopping. You don't finish what you start. The net force gives you acceleration, which will give you $v^2$. You need to keep going. Not just write one equation then stop.

 

[Drizzy]

How am I supposed to keep going when I don't understand. Can you explain the equation that is marked with the number 2 ?

 

[PeroK]

How am I supposed to keep going when I don't understand. Can you explain the equation that is marked with the number 2 ?

First, you wrote:

$F1 = sin(30)mg$

And just stop.

But, sin(30) = 1/2, so $F1 = mg/2$ That's not complicated!

Then you wrote:

F1 - force of friction = ma

And just stop. With 0.85F = force of friction, you have:

mg/2 - 0.85F = ma ...

As I suggested above, I think this problem is too long and complicated. I think you're not used to putting several ideas together. That said, it's not an easy problem and you can see by all those arrows that your teacher got into a bit of tangle trying to solve it!

Maybe someone else can step in, but I think I've shot my bolt on this.