Find acceleleration of object with angle only.

[dupdup]

Calculate the acceleration of a skier heading down a 10 degree slope assuming the coefficient of friction for waxed wood on wet snow.

I have drawn out the free body diagram but I am lost as how to find this without the persons mass. Is the question saying friction is the coefficient of waxed wood on wet snow (it is .1 for kinetic friction) or do i have to plug it into the equation f=uFnormal?

 

[Yuqing]

If you've drawn the free body diagram, you should be able to write a force equation for the skier. Now remember, you are trying to find acceleration, not force, so the masses on both sides of the equation should cancel.

 

[dupdup]

Im sorry but what force equation are you talking about? I know that if there was no friction i would use a=gsin(theta) but they are saying there is friction.

 

[dupdup]

Would weight and the ground cancel each other out since weight is pushing on the ground but the ground is pushing back? that would leave the kinetic friction and acceleration?

 

[Yuqing]

You don't really need to consider the direction perpendicular to the slope, the forces cancel exactly. Consider the forces parallel to the slope (i.e. gravity and friction). The acceleration of the skier is called by the sum of those two forces.

 

[dupdup]

Ok so friction is just .1N and gravity is 9.8m/s^2 so i would add .1 to it making acceleration 9.9m/s^2? Or is it .1 times 9.8 making it .98 m/s^2? The second one sounds right to me since the angle is only 10 degrees and there is friction.

 

[Yuqing]

0.1 should be the coefficient of friction and not the actual frictional force (what does your problem say?). Also, the normal force isn't simply g, its the direction of weight perpendicular to the slope.

 

[dupdup]

Wow i guess that's why they say read the whole problem in part b it says refer to question 38 and it gives an equation to use a=g(sin(theta)-ukcos(theta).