Posted by **jane** on Thursday, June 18, 2009 at 8:53am.

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?

- help with physics please! -
**drwls**, Thursday, June 18, 2009 at 10:07am
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.

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