Kinetic friction and finding acceleration?

Question

A 42kg block of ice sliding down a 34* plane has a kinetic coefficient of .06, what is the acceleration of the block down the incline?

Answer 1

So if you rotate the coordinate system, this problem is TEN TIMES easier, assuming I did it correctly...

So we have:

Mk= .06 -- coefficient of kinetic friction

mass=m=42 kg

we want to find: acceleration=a

Fg= force of gravity

N=normal force

Ff=frictional force

when you set up your y component equation it should look like this

N-Fg sin 34*=0=Fnet(y component)

* I think its 0 because you rotated the coordinate system so that the direction of the object moving is in the x component direction if that makes sense. And it it is not moving up or down so it is stationary and the acceleration=0 here there for the Fnet does too

N=Fg sin 34*

NOTE: Fg=42kg x 9.8m/s^2=411.6

N=411.6 x sin 34

N=230.1638 N

NOW DEALING WITH THE X COMPONENT:

Ff- Fg cos34= ma=Fnet(x component)

Ff= N x Mk

   = 13.8098

13.8098 - (411.6(the Fg, remember?)x cos34) = 42 kg x a

-327.42203 N = 42kg x a

ANSWER:

a=-7.796 m/s^2

I hope that's right :)