# Finding the distance between two points in terms of variables

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1. Oct 30, 2014

### dapiridoob

1. A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A block of inertia m is pressed against the free end of the spring until the spring is compressed a distance d from its relaxed length. Call this position A. The block is then released and moves up the ramp until coming to rest at position B. The surface is rough from position A for a distance 2d up the ramp, and over this distance the coefficient of kinetic friction for the two surfaces is μ . Friction is negligible elsewhere.

2.
What is the distance from A to B? Suppose the values of μ, k, d, θ, and m are such that the spring fully extends, but the block never goes higher than 2d .
Express your answer in terms of some or all of the variables k, μ, m, d, and θ

3. First I started by solving for the change in x.
I got to the point where deltax=v^2/( μgcos(θ)+gsin(θ)) through kinematics and force equations. I'm very certain that this equation is the correct way to find the distance a block moves up a ramp with friction. Then, I replaced delta x with 1/2k(2d)^2 and solved for d.

I ended up d=the square root of (2V^2/(( μgcos(θ)+gsin(θ))k))/2.

This answer does not seem to be correct and there seems to be a lapse in logic in how I'm going about this.

2. Oct 30, 2014

### BvU

Hello Dap, and welcome to PF :)

Use of the template can be improved: sorting out relevant equations under 2) would help a lot here!
And all variables listed and explained under 1).
The problem formulation is also weird. "the block never goes higher than 2d" is paart of that -- and it makes me wonder if the "Friction is negligible elsewhere" is just there to confuse you, or perhaps the exact wording of the exercise is different ?

I don't know what v is. Do you ? I think this comes from another context. I'm very certain that it's not applicable here.

You bring in a delta x which is probably the distance between A and B that the exercise asks for ?

The dimensions of deltax=v^2/( μgcos(θ)+gsin(θ)) are correct (energy/force = length) though. Perhaps you want to use some other energy instead of ½mv2 ?

For delta x = 1/2k(2d)^2 I am very certain that is completely wrong. Where does that come from ? Why the 2d ?