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Tanero
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Homework Statement
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
Homework 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
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.
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