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A particle of mass $m$ is attached between two identical springs on a horizontal frictionless tabletop. The springs have spring constant $k$, and each is initially unstressed. (a) If the mass is pulled a distance $x$ along a direction perpendicular to the initial configuration of the springs, as in Figure $\mathrm{P} 8.47$, show that the potential energy of the system is $$ U(x)=k x^{2}+2 k L\left(L-\sqrt{x^{2}+L^{2}}\right) $$ (Hint: See Problem 66 in Chapter 7.) (b) Make a plot of $U(x)$ versus $x$ and identify all equilibrium points. Assume that $L=1.20 \mathrm{~m}$ and $k=40.0 \mathrm{~N} / \mathrm{m} .(\mathrm{c})$ If the mass is pulled $0.500 \mathrm{~m}$ to the right and then released, what is its speed when it reaches the equilibrium point $x=0 ?$

If this distance is X this is L then that distances elsewhere to us X square on things distances Tetteh So we need to know what is the highest under the comfort of the force ethics. So FX equals if who signed Tetteh on f is que down to the extension and go Santa is eso You will come out to the negative zero to x There are two off the springs so two f x t x equals negative toe integration zero to eggs K x minus que el X over square root off elsewhere plus X square in traditional p x and this will come out to be gay X square That's what the blast site now K X squared plus to K l l minus experts us l squared. Yeah, putting the values in, we would have Yeah, the U versus X. We look like this on DFO or the last part. We would have initial kinetic energy plus initial potential energy equals have times mass 1.18 kg times the square we final square So we finally will come out with 0.8 to 3 meters per second

Let's discuss the set up for this problem were given the position of the mass that suspended from a spring, it's going to be why equals why not? Okay at times the code sign of tea route K over em. So we know that K. Is going to be the stiffness of the spring. And we want to find Dy DT ross has to find other values. Um so we see that if we look for Dy DT, we're also going to want to eventually find um the second derivative as well. So what this is gonna look like is Dy DT will be um taking why not? Than the co sign that's going to change to a negative sign of this whole thing. And then we're gonna multiply that by the um derivative of this internal component. But we see that this internal component is just a constant right here with the variable right here. So we'd end up just multiplying it by this right here. And then we do the same thing when calculating the second derivative

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