Friction Exercise

In my solution, I will not explain all the steps.
$F_{min}x=\frac{1}{2}kx^{2}+150x$
As we know ,to move $\textcolor{#20A900}{kx=25}$
$150+\frac{1}{2}kx =F_{min}$
$\textcolor{#3D99F6}{\boxed{F_{min}= 162.5}}$

The friction forces for block $A$ and block $B$ are $150$ and $25$ , respectively. My approach consists of a "virtual laboratory". The process is:

1) Try different values of $F$ , starting with something slightly greater than $150$ .

2) For each $F$ , run the time domain simulation until block $A$ stops (starts to move backward after initially moving forward). This is the point at which the spring force is greatest

3) Compare the final spring force against the friction force for block $B$ . The minimum required value of $F$ is the value of $F$ for which the final spring force is equal to the friction force for block $B$ . This minimum value comes out to be $162.5$ .

Code is attached.

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import math

# Constants

mA = 20.0
mB = 10.0
uA = 0.75
uB = 0.25
g = 10.0
k = 5.0

dt = 10.0**(-5.0)

###################

# Friction forces

FAf = uA*mA*g
FBf = uB*mB*g

print FAf
print FBf
print ""

#>>> 
#150.0
#25.0

###################

# Trial value of F

F = 162.5

###################

# Time-domain simulation

t = 0.0

xA = 0.0
xAd = 0.0
xAdd = 0.0

while xAd >= 0.0:

    xA = xA + xAd*dt
    xAd = xAd + xAdd*dt

    FA = F - FAf - k*xA

    xAdd = FA/mA

    t = t + dt

###################

# When block A comes to a stop.....
# compare final spring force against block B friction force

print dt
print t
print ""
print F
print (k*xA/FBf)

#1e-05
#6.28319999993

#162.5
#1.00000392701
#>>>