# How to quantify the resistive force of a spring?

Kymani Hatfield 2022-10-14 Answered
The situation: I have an ideal spring, at equilibrium, sitting on a table. I then put a weight onto the spring which compresses the spring.
The spring has a height of 1m and spring constant k. The weight has a mass, m. The weight compresses the spring by 0.5m. If we take the table height to be the zero of gravitational potential energy, then the work done on to the spring, by the weight, is:
$W=-F\cdot \mathrm{\Delta }h=mg\left(0.5\phantom{\rule{mediummathspace}{0ex}}\text{m}-1\phantom{\rule{mediummathspace}{0ex}}\text{m}\right)=\frac{1}{2}mg$
My question is: What is the work done on to the weight by the spring?
If we keep the conventions the same (spring is system, weight is surroundings), then the spring should be doing negative resistive work on to the weight while it's lowering.
In other words, shouldn't the work actually b
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Phoebe Medina
The work done on the mass by the spring is equal to:

where X is the distortion of the spring. The force on the mass is $\stackrel{\to }{F}=-k\stackrel{\to }{X}$ , and $\stackrel{\to }{X}\cdot \mathrm{d}\stackrel{\to }{X}=x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}X.$
For your problem, the initial distortion is zero (${X}_{1}=0$) and ${X}_{2}=0.5$ m. The work done is
$W={\int }_{0}^{0.5}\left(-kX\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}X={\frac{-k{X}^{2}}{2}|}_{0}^{0.5}=\frac{-k}{8}.$
The work by the spring on the mass is, indeed, negative.
Lagniappe: If the mass starts with zero velocity at the top and ends with zero velocity at the 0.5 m point, the net work should be zero, so we should get
$\frac{-k}{8}+\frac{1}{2}mg=0$
$\frac{1}{2}mg=\frac{k}{8}$
The equilibrium position (the position at which the mass would continually rest on the spring) is going to be

halfway between the top point and the 0.5 m point in your problem. That's what we expect for a spring-mass-gravity oscillator.
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pezgirl79u
I'm not sure I follow where you decide to designate the zero of potential energy at the table height as this is where the potential energy of the spring is at its maximum as it is fully compressed. It seems counter-intuitive to me to want to set this as a zero (unless you're talking about a gravitational zero). Nonetheless, I shall try to answer.
There is a convention that work is gained from a loss of potential energy, in this case gravitational potential energy. You could say that the spring doesn't do any work on the weight because the spring's potential energy is increasing or you could say the spring is doing negative work. The weight's potential energy due to gravity is decreasing (its height with respect to the Earth/table decreases). Thus, according to this setup, gravity is doing positive work (via the weight) and the spring is doing negative work (via "resistive force"). The spring's elastic potential energy increases and the weight's gravitational potential energy decreases.
Work is indeed the net force times the displacement (dot product, actually) but whether or not you choose gravity or the spring or the weight depends on whose potential you care about and what sign/direction convention you adopt.