PE at A = 3mgx
WD = Fs
KE = 1/2mv^2
The Attempt at a Solution
The question I am stuck on is part ii.
I worked out from part i that the PE at A is 3mgx, so therefore all this must go towards the KE and the sound, and doing work against friction, etc...
I said that the change in energy must equal the work done so -3mgx = μmgd, and the mg must cancel so d = -3x/μ
However the mark scheme says d = 3x/μ, without the negative. So have I done something wrong or is this just to do with the direction I have chosen?