- #1
ThePiGeek314
- 10
- 0
Homework Statement
Calculate the deceleration of a snowboarder going up a 5.0° slope, assuming the coefficient of friction for waxed wood on wet snow ( = 0.1).
Homework Equations
F = ma
W = mg
Fs = (Fn)
Fk = k (Fn)
The Attempt at a Solution
Answer key says it's 1.83 m/s2. I fiddled around with the numbers I got and I was able to get the right answer, but I don't understand why it's the right answer. Below is my work.
Normal force = 9.8 (assuming the snowboarder has a mass of 1 kg)
Force pushing the snowboarder UP the hill: 9.8 sin (5 degrees) = 0.854 (Is this right?? Is there a force pushing him up the hill??)
Kinetic friction = * Fn = 0.1 * 9.8 = 0.98 N
Sum the horizontal forces, applying Newton's Second Law of Motion...
a = (∑Fx) / m
(a = acceleration, Fx = all horizontal forces, m = mass)
Because the 0.854 is the force pushing him up the hill, and the kinetic friction works opposite to that, I subtracted the kinetic friction value from the 0.854. So...
a = (0.854 - 0.98) / m......(m is assumed to be 1)
a = -0.126 / 1
a = -0.126 m/s2
But if I simply add the two force values that I found - add the kinetic friction and the force parallel to the hill -- I get 1.834, which is the correct answer. That means that my assumption that 0.854 N was pushing the snowboarder up the hill was wrong. So both the 0.854 N and the kinetic friction must be working in the same direction.
But if there's no force pushing him up the hill, and I'm not given an initial velocity up the hill, then how can there be any motion at all? He'd be going down the hill, not up it.
Please explain why the 0.854 N and kinetic friction are working in the same direction, since kinetic friction is supposed to work in the opposite direction of the attempted motion.