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Feb
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Jan
13
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awarded  Popular Question
May
10
comment Dropping a weight onto a spring scale
@Rations: When you gently put a mass $M$ just above the scale, and drop it from there, the scale initially oscillates between $0$ and $2M$. It is only after damping (and some loss of energy) that you get a reading of $M$.
May
9
comment Dropping a weight onto a spring scale
@Rations: $R\;g\;=\;k\;x$ is only about the scale; at any moment, the scale's reading $R$ is proportional to the displacement of the scale $x$, and the combination of the scale's spring and mechanism is such that this equation holds, so that the scale's reading (at equilibrium) gives the mass of what's on it. $g$ is the gravity of earth assumed by the scale's manufacturer (or calibration).
Apr
29
revised Dropping a weight onto a spring scale
Revise kind of scale perhaps allowing the method to be used
Apr
29
revised Dropping a weight onto a spring scale
We won't have time to make the reading, and it would far off-scale or/and useless
Apr
29
revised Dropping a weight onto a spring scale
Actually answer
Apr
28
revised Dropping a weight onto a spring scale
Polish comment on relative error on k
Apr
28
revised Dropping a weight onto a spring scale
Discuss relative error
Apr
28
revised Dropping a weight onto a spring scale
Discuss the case of a spring soft enough that the method can work
Apr
27
revised Dropping a weight onto a spring scale
The method is doomed
Apr
27
revised Dropping a weight onto a spring scale
Mention we should not trust the result
Apr
27
revised Dropping a weight onto a spring scale
Explain the case h=0
Apr
27
revised Dropping a weight onto a spring scale
Give hypothesis
Apr
27
revised Dropping a weight onto a spring scale
Expand with numeric values and comparison to other answer giving an approximation
Apr
27
awarded  Teacher
Apr
27
answered Dropping a weight onto a spring scale
Apr
27
comment Dropping a weight onto a spring scale
Correction: if we make $h=0$, we should obtain $m=2M$.
Apr
27
comment Dropping a weight onto a spring scale
If we make $h=0$, the displayed mass$$m=\sqrt{2\;k\;h\;M\over g}$$ is $m=0$, when it should be $m=M$ (what is displayed when we delicately put the weight on the scale, as we should). Also this solution does not account for the statement's fact that the display $m$ is more than $M$ for all $h>0$. So this can't be exactly right. Perhaps it is a valid approximation in some unstated domain, like $$h\gg{M\;g\over k}$$PS: what a pain that the notation makes $m>M$.