A skier has just begun descending a 20 deg slope. Assuming that the coefficient of kinetic friction is 0.10, calculate the acceleration of the skier, and his final velocity after 8.0s.
thanks
Sketch a Free-Body Diagram. I have a +x direction down the plain and a +y direction upward and perpendicular to the plain. There is a weight, w, on the skier directed straight down, there is a normal force, N, on the skier directed in the +y direction, there is a friction force, f, acting in the -x direction. The inclination of the plain from the horizontal is $\displaystyle \theta$.
According to Sir Newton:
Since the skier doesn't accelerate in the y direction, $\displaystyle a_y = 0$:
$\displaystyle \sum F_y = N - w~cos(\theta) = 0$
$\displaystyle N = w~cos(\theta) = mg~cos(\theta)$
where m is the mass of the skier.
And
$\displaystyle \sum F_x = -f + w~sin(\theta) = ma$
$\displaystyle -\mu N + mg~sin(\theta) = ma$
We know that $\displaystyle N = mg~cos(\theta)$ from the y equation, so
$\displaystyle -\mu (mg~cos(\theta)) + mg~sin(\theta) = ma$
where $\displaystyle \mu$ is the coefficient of kinetic friction.
We can divide through by the common m:
$\displaystyle a = g~sin(\theta) - \mu g~cos(\theta)$
As far as the velocity at t = 8.0 s, the acceleration is constant and you know the skier started from rest, so simply apply $\displaystyle v = v_0 + at$.
-Dan
I'm not sure if it'll make a difference in the way it's explained or how the answer works out with the mathematical sign that is in front the first and second line...if it does can someone help me change it cause I think it's from calculus but I haven't learned that yet
There is no Calculus in my solution.
Remember that forces are vector quantities. I am using the usual shorthand for vector components: When I say, for example, the friction force is in the -x direction and use -f for that quantity, I am actually using the x component of the friction force, which has a magnitude of the size of the friction force and contains a negative sign because it is pointing in the -x direction. The same argument holds for the y component of the weight.
-Dan