2.50 kg particle (the function U(x) has the form bx2 and the vertical axis scale is set by Us = 2.0 J).(a) If the particle passes through the equilibrium position with a velocity of 55.0 cm/s, will it be turned back before it reaches x = 18.0 cm?(b) If yes, at what position (in cm), and if no, what is the speed of the particle (in cm/s) at x = 18.0 cm?

Question

drwls · Answer

I don't know what you mean by Us = 2.0 J
This looks like a mass on a spring problem for which conservation of energy should be used. The sum of potential and kinetic energy equals a constant. At the equilibrium position, the kinetic energy is
(1/2)M V^2 = (1/2)(2.5)(0.55)^ = 2
= 0.378 J
Yhat equals the consnta total energy of the system.
You need to determine the spring constant somehow so you can calculate the potential energy when x = 0.17 m. From that, you can get the speed.