# 43 a derive an expression for the potential energy of

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Unformatted text preview: b = 2 N/m3, taking U = 0 <t.x = O. (b) Graph the potential energy curve for 0, and use it to find the turning points for an !;>ject hose total energy is -1 J. w :> >1on The equation U(x)- E = 0 is equivalent to x4 2(afb)x2 ~ 4(E/b) = O.The quadratic formula gives X = :I:{(a/b) :I: [(alb? + 4(Elb)]l/2P/2 for Umin < E < 0, and x = :f:{(a/b) + [(a/b)2 + 4(Efb)F/2P/2 for E >0. For the particular values given (E = -1 J), the positive turning points are x = V(5:f: .Jfi)/2 m = 0.662 m and 2.14 m, as can be seen in the graph. 'he force is conservative (anyone-dimensional ~~en by an integrable function of position is), so tential energy can be found from Equation 8-2a: ~ - ',;;(z - U(Xl) 'ii. A< f (~dz bx') 2 2 Xl) Section 8~5: Force and Potential Energy Note: In the following problems, motion is restricted to one dimension. = -"2(X2 - a + 4(X2 b 4 - 4 Xl)' Problem 44. Figure 8-36 shows the potential energy curve for a certain particle. Find the force on the particle at each of the curve segments shown. 3 (b) ~fine t...
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## This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.

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