Oscillation on inline

Let $x$ denote the block's position on the incline relative to its starting position with positive values denoting positions down the wedge. In the $x$ -direction, the block experiences three forces: a component of its weight $Mg \sin \theta$ , the tension in the spring $-kx$ , and friction, which has magnitude $\mu Mg \cos \theta$ and initially acts in the same direction as the spring. Let $x_1$ denote the block's position at the bottom of its first oscillation. At $x = 0$ , the block has 0 velocity, and from $x = 0$ to $x = x_1$ , the net work of gravity, the spring, and friction is 0 because the block again has 0 velocity. This means $\int_0^{x_1} (Mg \sin \theta - \mu Mg \cos \theta - kx)\, dx = 0$ and thus $x_1 = \frac {2Mg}k(\sin \theta - \mu \cos \theta).$ At this point, the block's velocity and friction change directions. Let $x_2$ denote the position of the block at the top of the first oscillation. By the same argument as before, the work integral (with friction acting in the same direction as gravity) must again equal 0, so $\int_{x_1}^{x_2} (Mg \sin \theta + \mu Mg \cos \theta - kx)\, dx = 0 \implies x_2 = -x_1 + \frac {2Mg}k(\sin \theta + \mu \cos \theta).$ In general, if $x_n$ denotes the block's position the $n$ th time it comes to rest, then $x_{n + 1} = -x_n + \frac {2Mg}k \big[\sin \theta + (-1)^{n + 1} \mu \cos \theta \big].$ This recurrence relation continues to hold until the block comes to rest at some $x_n$ where the magnitude of the combined force from gravity and the spring, $|Mg \sin \theta - kx_n|$ , is less than or equal to the magnitude of the friction force, $\mu Mg \cos \theta$ . When this happens, gravity and the spring can no longer overcome friction, so the block remains completely at rest. In this particular case, this happens at $x_3$ . Taking $x_0 = 0$ and applying the recurrence relation three times gives the values $x_1 = \frac {2Mg}k(\sin \theta - \mu \cos \theta) = \SI{0.12}{m}, \\[0.3em] x_2 = -x_1 + \frac {2Mg}k(\sin \theta + \mu \cos \theta) = \SI{0.04}{m}, \\[0.3em] x_3 = -x_2 + \frac {2Mg}k(\sin \theta - \mu \cos \theta) = \SI{0.08}{m}.$ Therefore, the block's final position is $x_3 = \SI{0.08}{m}$ from its starting position, where the spring is stretched by $\boxed{\SI{8}{cm}}$ .

Since the solution directly below mine is slightly flawed, I have posted a code solution just to make sure.

So, the gravitational force acting on the block in the ramp's direction is:

$\displaystyle \bold{F}_g = mg \sin(\theta)$

Spring force:

$\displaystyle \bold{F}_s = -k(x-l)$

Friction force:

$\displaystyle \bold{F}_{\mu} = \pm \mu mg \cos(\theta)$

The friction force depends on the direction of velocity. Programming all these forces into a Python simulation program, and calculating net force at each point, thereby calculating the acceleration, we can numerically integrate.

Here's the code:

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import math

time = 0
deltaT = 10**-6 #timestep

#position and velocity
x = 10
xDot = 0


#constants
l = 10
k = 100
m = 1
mu = 1/(math.sqrt(51))
g = 10
theta = math.asin(7/10)



while time <= 6:   #Enough time to get block at rest
  Fg = m*g*math.sin(theta)   #gravitational force
  Fspring = -k*(x-l)      #spring force

  #Frictional force depends on direction of velocity
  if xDot > 0:
    frictionForce = -mu*m*g*math.cos(theta)
  else:
    frictionForce = mu*m*g*math.cos(theta)

  totalForce = Fg + Fspring + frictionForce #Total force

  #Numerical Integration
  xDotDot = totalForce/m
  xDot += xDotDot*deltaT
  x += xDot*deltaT


  #Updating time value
  time += deltaT



print("At rest, position = " +str(x))

As we can see, extending the time, the final elongation appears to be $8$ centimeters.

As the block moves down the incline, the spring starts getting elongated.

At maximum elongation, the spring shall have the tendency to move the block upwards ( or towards the spring's equilibrium position). We know force component mg sin theta shall act downwards the incline. Now since the block has tendency to move upwards, the friction force shall be acting downwards the incline.

Hence, kx = mg sin theta + f ; f is the force of friction at that instant.

Therefore, f = kx - mg sin theta

We know f is smaller than equal to its limiting value, which is equal to umg cos theta for an inclined plane situation.

OR kx - mg sin theta = f ( instant ) <= f ( limiting ) <= u mg cos theta

OR

kx <= u mg cos theta + mg sin theta

Solve the equation to obtain the maximum elongation.