particleof mass m = 1.18 kg is attached between two identical springs on ahorizontal frictionless tabletop. The springs have spring constantk, and each is initially unstressed. (a) If the mass of pulled adistance x along a direction perpendicular to the initialconfiguration of the springs, show that the potential energy of thesystem is: U(x)=K(X^2) + 2KL (L- square-root(X^2 + L^2) {look above thequestion for better viewing equation } (b) Make a plot of U(x) versus x and identify all equilibriumpoints. Assume that L = 1.20 m and k = 40.0 N/m. (c) If the mass is pulled 0.500 to the right and then released, what is its speed whenit reaches the equilibrium point x = 0? (d) Using the potentialenergy of the system find the force as a function of x.
NOTE: the picture is top view pictute
U(x) = kx^2 + 2kL(L - rootx^2 + L^2)particle of mass m = 1.18 kg 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 of pulled a distance x along a direction perpendicular to the initial configuration of the springs, show that the potential energy of the system is: U(x)=K(X^2) + 2KL (L- square-root(X^2 + L^2) {look above the question for better viewing equation } (b) Make a plot of U(x) versus x and identify all equilibrium points. Assume that L = 1.20 m and k = 40.0 N/m. (c) If the mass is pulled 0.500 to the right and then released, what is its speed when it reaches the equilibrium point x = 0? (d) Using the potential energy of the system find the force as a function of x.NOTE: the picture is top view pictute
