The max compression of a spring with friction?

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

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