- #1
Anon2459
Homework Statement
A mass of m = 1.40 kg is placed on a 2.00 m ramp. The angle of the ramp can be adjusted by changing the height of the top of the ramp.
In reality there is a small amount of friction between the block and the ramp: μs = 0.261, μk = 0.119.
If the ramp (with friction acting) is now lifted so that θ = 48.0o and the block now slides from rest the full 2.00 m down the ramp what is its speed at the bottom?
Homework Equations
V^2 = -2 x g x (h-ugcosangle x d)
height of ramp = 0.7020408163
The Attempt at a Solution
V^2 = -2 x 9.8 x (-0.7020408163 + 0.119 x cos48 x 2) = wrong answer
Which friction do i use? Static or kinetic? Or is my approach wrong entirely?
Not sure where i went wrong[/B]