Spring whose spring constant is a piecewise function of displacement—is this possible?

A bungee cord is essentially a very long rubber band that can stretch up to four times its unstretched length. However, its spring constant varies over its stretch [see Menz, P.G. “The Physics of Bungee Jumping.” The Physics Teacher (November 1993) 31: 483-487]. Take the length of the cord to be along the $$x$$-direction and define the stretch $$x$$ as the length of the cord $$l$$ minus its unstretched length $$l_0$$; that is, $$x=l−l_0$$. Suppose a particular bungee cord has a spring constant, for $$0≤x≤4.88$$ m, of $$k_1=204$$ N/m and for $$4.88 \mathrm{m}≤x$$, of $$k_2=111$$N/m. (Recall that the spring constant is the slope of the force $$F(x)$$ versus its stretch $$x$$.) (a) What is the tension in the cord when the stretch is $$16.7$$ m (the maximum desired for a given jump)? (b) How much work must be done against the elastic force of the bungee cord to stretch it $$16.7$$ m?

The given answer is (a) 2.32 kN, (b) 22.0 kJ.

I'm struggling to wrap my head around the idea of a bungee cord with these properties. Suppose I attach a weight to the cord having mass slightly less than $$204 \cdot 4.88 /g = 101.5$$ kg. Then the cord will hang in equilibrium at just above $$x=4.88$$ m below the unstretched length.

Then, suppose I attach a feather to the weight. Now it will sag past $$4.88$$ m, and the spring constant instantly changes (!) to $$111$$ N/m, putting the new equilibrium at $$101.5 g/111 = 8.97$$ m. Am I correct in understanding that the weight would suddenly fall $$4$$ m? And if I then removed the feather, would the equilibrium point remain at about $$8.97$$ m, even though its weight has been restored to what it was before (when the equilibrium was at $$4.88$$ m)? This seems very counterintuitive to me.

Ignoring my intuition, I proceeded under the assumption that $$x=16.7$$ (as given in the question) is in the $$k=111$$ domain. Then the magnitude of the tension is $$kx = 1854$$ N, which doesn't match the answer given for (a).

In (b), I noticed that the spring force $$kx$$ is a piecewise function of $$x$$, and integrated like so:

$$\int_0^{4.88} 204 x\, dx + \int_{4.88}^{16.70} 111x\,dx = 16586 ~\mathrm{J}$$

Swing and a miss.

In both cases, I have the correct order of magnitude, so my best guess is that I'm forgetting to account for gravity somehow. But we aren't given the mass of the the bungee jumper, so I don't see how I would incorporate an $$mg$$ term in here.

Edit: I think I have a partial guess for what's going on in (a). If we redefine the tension force as

$$F_s (x) = \int_0^x k(x')\, dx'$$

Then

$$F_s (16.7) = \int_0^{4.88} 204 \, dx + \int_{4.88}^{16.70} 111\,dx = 2308 ~\mathrm{N} \approx 2.32~\mathrm{kN}$$

To me, this looks like the sum of the tension in two different springs, each having two different spring constants and stretch values. Is this correct?

This is one of those questions that are easier to grasp using a graphical view rather than calculus notation, at least for me. The key here was the hint that the spring constant is the slope of the force vs. stretch graph. Here is the force vs. stretch graph:

As one can see, the force doesn't fall back to $$0$$ at $$4.88$$ meter. It continues to rise at the rate of $$111 ~ N/m$$, but it still has whatever tension was necessary to reach $$4.88 ~ m$$. So, at $$16.7 ~ m$$ the tension is the sum of the force necessary to stretch to $$4.88 ~ m$$, plus the additional force necessary to go from $$4.88$$ to $$16.7 ~ m$$. That is just what you found: $$204 ~ N/m \cdot 4.88 ~ m + \left( 16.7 m - 4.88 m \right) \cdot 111 ~ N/m = 2300 ~ N$$.

Using the area under the graph, we can also see that the work done against the elastic force has three pieces: the triangle from $$0$$ to $$4.88 ~ m$$, the triangle from $$4.88 ~ m$$ to $$16.7 ~ m$$, and the rectangle from $$4.88 ~ m$$ to $$16.7 ~ m$$. We have $$W_{1} = 0.5 \cdot \left(204 \cdot 4.88 \right) \cdot 4.88 = 2429 ~ J$$. $$W_{2} = 0.5 \cdot \left(111 \cdot \left( 16.7 - 4.88 \right) \right) \cdot \left( 16.7 - 4.88 \right) = 7754 ~ J$$. $$W_{3} = \left( 204 \cdot 4.88 \right) \cdot \left( 16.7 - 4.88 \right) = 11767 ~ J$$.

The total work is $$W = 21950 ~ J \approx 22 ~ kJ$$.

Your question edit is correct. I can understand your confusion. You integrated incorrectly the first time.

$$k = \frac{\mathrm d F}{\mathrm d x} \quad \Rightarrow \quad F(x) = \int_0^x k(x') \,\mathrm d x' \; .$$

The spring constant $$k(x')$$ is a piecewise function as you say. It is $$k_1$$ for $$x' < 4.88 \mathrm m$$, and $$k_2$$ above that.

You can think of the 'factor' $$\mathrm d x$$ as being the multiplication by $$x$$ which gives the force. ($$F = k x$$ is the equation for an ordinary spring. The $$x$$ gets replaced with $$\mathrm d x$$ here, with each $$\mathrm d x$$ being 'weighted' by the spring constant that applies at the current value of $$x$$).

Hope that helps.

$$F_1=k_1\,x_1$$

$$F_2=F_1+k_2\,(x_2-x_1)$$

you obtain $$~F_2=2307 [N]~$$

the work is the "area" below the curve $$~F(x)~$$. $$~(W=\int_{x_i}^{x_f} F(x)\,dx)~$$
$$W_1=\frac 12 F_1\,x_1\\ W_2=\frac 12 (F_2-F_1)\,(x_2-x_1)+(F_1)\,(x_2-x_1)$$

the total work is $$~W=W_1+W_2=21950 [N\,m]$$