1. The problem statement, all variables and given/known data A small block of mass m=1 kg and v0=6.32m/s hits a massless relaxed spring with spring constant k= 3 N/m, which starts to be compressed as the block continues to move horizontally. There is friction between the block and the horizontal surface, and it is not uniform. As a function of distance, the friction coefficient varies like μ(x)=αx, with α= 0.7 m^−1. For simplicity that static and dynamic friction coefficients are the same. What time t1 does it take for the block to travel between x=0 (relaxed spring) and x=x1 (block at first stop)? (in seconds) https://courses.edx.org/static/content-mit-801x~2013_SOND/html/final_p06.png [Broken] 2. Relevant equations I chose x axis positive to the right. x0=0 at point of contact, x1=first momentarily stop Ffriction=-μmg=-αxmg Fspring=-kx Newton's Second Law ∑F=m*d2x/dt2 3. The attempt at a solution -kx-αxmg=m*d2x/dt2 -(k+αmg)x=m*d2x/dt2 new konstant K=k+αmg -K/m=d2x/dt2 This is SHO with general solution x(t)=Acos(ωt+θ) ω=√(K/m)=3.16 To find A, and θ our initial conditions are: x0=x(t=0)=0 v0=v(t=0)=6.32 m/s Thus, 0=x(t=0)=Acos(θ) This is true when θ=pi/2 or -3/2pi . Question: which angle should I use ??? v0=v(t=0)=6.32/-ω=Asin(θ) A=√(x(t=o)^2+(v(t=0)/ω)^2)=√(0+(6.32/3.16)^2/)=2 So, I know that my amplitude is 2, x1=2m. (Trying with 3/2pi. pi/2 can't feet because I am getting negative time.) 2=x(t1)=2*cos(3.16t-3/2pi) 1=cos(3.16t-3/2pi) This is true when (3.16t-3/2pi)=0 t1=1.49 sec. Please, check my solution if it seemed to be right. And my t1 should be also half a period T=pi/ω=pi/3.16=0.99s. Isn't it? I am confused.