Solving a Mass-Spring System: Max. Displacement of Mass

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SUMMARY

The discussion focuses on solving a mass-spring system involving a 1.2 kg block attached to a spring with a force constant of 12.0 N/m and a coefficient of kinetic friction of 0.15. The user successfully derived the speed of the mass as a function of displacement, calculated the maximum speed of 2.06 m/s at x = 0.147 m, and determined the spring compression of 0.505 m when the mass momentarily stops. However, the user struggles with deriving an expression for maximum displacement after multiple oscillations, indicating a need for further clarification on energy conservation principles in oscillatory motion.

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  • Understanding of Newton's laws of motion
  • Familiarity with Hooke's Law and spring constants
  • Knowledge of conservation of energy principles
  • Basic calculus for deriving functions and solving equations
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Homework Statement


A block of mass 1.2kg, resting on a horizontal surface of coefficient of kinetic friction of .15, is attached to a linear spring of force constant 12.0 N/m. When the spring is unstretched the mass is at a position x=0. THe mass is pulled to the right streching the spring a distance x=0.8 m, and relased from rest at t=0. The mass accelerates the the left and momentarily comes to rest at point B.

a) Derive an expression for the speed of the mass as a function of x.
b)Calculate the maximum speed of the mass during this time. At what position x does this maximum speed occur?
c)Calculate the distance the spring is compressed whne the mass stops momentarily at point B.
d)Derive an expression for the maximum displacement of the mass at the end of each half oscillation as a function of hte number of half oscillations, n.


Homework Equations


Conservation of energy is what i used for a, b, and c. No idea how to do d.


The Attempt at a Solution



a)
.5*k*.8^2-.5*k*x^2-m*g*.15*(.8-x)=.5*m*v^2
k=12, m=1.2, g =9.8
sqrt(2(6*.8^2-6*x^2-1.764(.8-x))))=v
which simplifies to
v=Sqrt[-10x^2+2.94x+4.04]

b)
v=Sqrt[-10x^2+2.94x+4.04]
dv/dx = (-Sqrt[10]*(x-.147))/Sqrt[-x^2+.294+.404]
dv/dx = 0
x=.147
V(.147) = 2.06 m/s
therefore
2.06 m/s @ x=.147

c)
V=0
Sqrt[-10x^2+2.94x+4.04] = 0
x = -.505

.505 m compressed

d)

OK, I think I got the previous parts correct, but this part seriously breaks my brain. I simply do not know where to start. I tried to make a function like

1/2*k*X(n)^2-m*.15*g*(X(n)-X(n+1))=1/2*k*X(n+1)

but i don't know how to solve it. Many thanks in advance.
 
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ok i redid my equation to 1/2*k*((x(n))^2-mg(-1^n(x(n))-(-1^(n+1))((x(n+1))=1/2*k*x(n+1)^2
 

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