# Jij5 l j xm rjct 5

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Unformatted text preview: orce on a particle in one-dimensional motion, when the potential energy is a straight line segment, is the negative Qftheslope of U(x)j i.e., Fx= -dU/dx (Equation 8-8). (This is, of course, just the conservative force represented by the potential energy.) From Fig. 8-36, we can determine the negative of the slopes: (a) -3 J11.5 m = - 2 N; (b) OJ (c) -(~4 Jl~ m) =8Nj (d) -(-1 J/l m) = 1 N, (e) -4 Nfl m = -4 N, (f) O. Problem 45. A particle is trapped in a potential well described by U(x) = 2.6x2- 4, where U is in joules, and x is in meters. Find the force on the particle when it's at (a) x = 2.1 mj (b) x::::: 0 mj and (c) x = -1.4 m. Solution For one-dimensional motion, Equation 8-8 gives the force Fx = -dU/dx = -d/dx(1.6x2 - 4) :::::2(1.6)x = -3.2x, where Fx is in newtons for x in meters. Therefore, (a) Fx(2.1 m) = -(3.2)(2.1) N = -6.72.N (force in the negative x direction)j (b) Fx(O) = OJ (c) Fx(-lA m) = -(3.2)(-1.4) N = 4.48 N. Problem 43 Solution. 134 CHAPTER 8 the spring (not the stretch). Find (a) the equilibrium length of the spring and (b) 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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