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If you suppose that the scale works like a spring, which seems reasonable, then during standard use, the displacement $x$ of the scale is proportional to the mass $m$. The equilibrium relation is $$ mg=kx,\tag{1}$$ where $k$ is the stiffness of the spring. Assuming that, when you dropped a mass $M$ from a height $h$, all kinetic energy (which is equal to ...


1

This hopefully improves on an approximation made in the other answer, while using the same hypothesis (doubtful in practice) that the scale has no damping or friction no energy is lost when the mass impacts the scale the scale's moving mechanism has negligible mass compared to the mass dropped I use the same notation except for the maximum reading of the ...



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