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Physics question, help please….?

A block with mass 0.5kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2m. When the spring is released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The force constant k is 100N/m. What is the coefficient of kinetic friction μk between the block and the tabletop?

1 Answers

E
Everardo Feil

Feb 20, 2025

The spring potential will be converted to friction work ½kx² = (μmg)dμ = ½kx² / mgdμ = ½100(0.2)² / 0.5(9.8)1.0 μ = 0.40816…μ = 0.41I hope this helps. Please remember to select a “Best Answer” from among your results, It’s good karma as it helps keep the exchange in balance.... Show More

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