A block of mass m attached to a spring of force constant k rests on a rough horizontal floor. The coefficient of kinetic friction is μₖ. Initially the spring is compressed by x₀ and the block is released. It crosses the natural-length position and first comes to rest at an extension x₁ on the other side. Then:
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Question

A block of mass m attached to a spring of force constant k rests on a rough horizontal floor. The coefficient of kinetic friction is μₖ. Initially the spring is compressed by x₀ and the block is released. It crosses the natural-length position and first comes to rest at an extension x₁ on the other side. Then:

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G1 Physics · Accepted Answer

From the initial compressed position to the first turning point on the other side, the block travels distance x₀+x₁. Energy loss to friction is μₖmg(x₀+x₁). Thus ½kx₀² = ½kx₁² + μₖmg(x₀+x₁). Factorising: ½k(x₀-x₁)(x₀+x₁)=μₖmg(x₀+x₁). Hence x₀-x₁ = 2μₖmg/k, so x₁ = x₀ - 2μₖmg/k.