The max compression of a spring with friction?

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Potential energy minus the work of friction is equal to the change in kinetic energy. The kinetic
energy in turn causes an equal amount of change in spring energy.

∆U − ∆Wf = ∆K = ∆Sp
mgh − μmgdcosθ = ½kx²

The box first travels 2 m along the plane, but also the distance x, by which the spring is compressed,
therefore: d = 2+x => h = (2+x)sinθ
==>
mg(2+x)sinθ − μmg(2+x)cosθ = ½kx²
This rearranges into the following quadratic equation:
kx² − 2mgx(sin−μcosθ) − 4mg(sinθ−μcosθ) = 0

At this point we can see that since θ=45° and μ=0.5, then sinθ−μcosθ = 1/√8
=>
kx² − 2mgx/√8 − 4mg/√8 = 0
kx² − mgx/√2 − mg√2 = 0
Using the values for k, m and g, we get:
500x² − 100x/√2 − 100√2 = 0
x² − x/(5√2) − (√2)/5 = 0

By the pq-formula, x can be expressed as:

x = [ 1 ± √(1 + 40√2) ] / 10√2

x₁ = −0.4658
x₂ = 0.60722...Show more

work = variation of kinetic energy
m g h - u m g h / tan@ - 1/2 k x^2 = 0
x = {2 [m g h (1 - u / tan@)] / k}^1/2
x = {2[10*10*2(1 - 0.5 / tan45)] / 500}^1/2 = 0.632 m...Show more