Recent content by dzimme2

Do you get 0.1257?

After changing the sign of f_{k} to restate : F_{net}_{x} = F - mgsin\theta - f_{k} = 0; f_{k} = F - mgsin\theta μ_{k} F_{N} = F - mgsin\theta μ_{k} mgcos\theta = F - mgsin\theta μ_{k} = \frac{F - mgsin\theta}{mgcos\theta} μ_{k} = \frac{(10 N) - (2kg)(9.8 m/s^2)sin 22}{...

Ohh, I see. Could prove to be the error with other problems as well. I'll try it out and see what I get. Thanks again!

I was under the impression that the kinetic frictional force would oppose the block's ability to slide down the plane due to gravity. While the applied force is pushing the box up the plane and gravity is pulling the block down the plane, wouldn't the frictional force be directed up the slope...

Homework Statement A 2 kg. block is pulled up a 22° incline with a force of 10N parallel to the incline. What is μ_{}k if it moves up with constant speed? Homework Equations F_{}N = Normal Force f_{}k = μ_{}k F_{}N The Attempt at a Solution Forces perpendicular to plane...