When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.21 m/s. The mass stops a distance S2 = 2.1 m along the level part of the slide. The distance S1 = 1.14 m and the angle θ = 29.50°. Calculate the coefficient of kinetic friction for the mass on the surface.

Question

anonymous · Answer

So you know that the box could not have a constant velocity as it traveled because it comes to a stop. So you know the net force isn't zero. The force of friction is slightly greater than the force of gravity pushing it down the slope. You can solve for that acceleration using the equation $$V^2 = V _{0}^2 + 2a \delta x$$ Once you solve for the accelaration you know that that acceleration is coming from the force of friction so you can say that $$ma= \mu  N$$ where n is the normal force which  in the case of an incline plane is given by $$N= mgcos(\theta)$$ so you get $$ma = mu mgcos(\theta) $$ for some reason it doesn't want to write mu but thats what that is. The mass cancles and from there you can solve for the coefficient of static friction given all the knowns