6.4 The potential energy function for aparticle executing linear simpleharmonic motion is given by V(x) =kx /2, where k is the force constantof the oscillator. For k = 0.5 N m',V(x)the graph of V(x) versus x is shownin Fig. 6.12. Show that a particle oftotal energy 1 J moving under thispotential must 'turn back' when itreaches x = + 2 m.Fig. 6.12

force constant = K = 0.5 N/m total energy = E = 1 J

Answered: The potential energy function for a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A simple pendulum of lengthL=0.972m with a bob of mass m=45.8g is pulled aside untilthe bob is at a height L/4 above its equilibrium position. Thebob is then released. Find the speed of the bob as it passesthrough a point where its height is L/7 above its equilibriumposition. What is the angle the pendulum makes with thevertical at this point?
Problems 10 and 11: A nonlinear spring has a spring force function of: F =-2x* (x+2) (N) 10. What is the spring potential energy function? Answer 10: 11. The spring is positioned horizontally along a frictionless surface. A 2 kg mass is pushed against the spring. By what amount should the spring be compressed so that when the mass is released, the max velocity of the mass is 5 m/s? Answer 11:
what will be the turning points if the particle's total energy is doubled? express in cm...enter in ascending order...separate by comma
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A particle that can move along the x-axis is part of a system with potential energy U(x) = A x 2 − B x , where A and B are positive constants. (a) Make a sketch of U(x) vs x. (b) Where are the particle’s equilibrium positions? (You may neglect the “point” where x is very large) (c) Describe how a particle would move if given small displacements from the equilibrium positions. For each of the points you identified in Part A, identify if it is a point of stable or unstable equilibrium.
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A simple pendulum of length L is set to oscillate in simple harmonic motion. The bob gravitational potential energy is zero at its lowest vertical point. When the bob's height is at half its maximum, h = hmax/2, then its velocity is: 10) (a) v = /ghmax/2 (b) v = ghmax (c) v = /2gh,max (d)v = 2ghmax < o O
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k 3. A box (mass = 100 g) is initially connected to a compressed (x = 80 cm) spring (k = 100 N/m) at point A. It was released and started moving along the horizontal surface (μ = 0.2) until it moves up along the inclined surface (μ = 0.3). The box then stops at point D alone the incline. Consider that e = 30 m and 0 = 30°. (a) What is the velocity of the box at point B?) (Ans: 25.24 m/s) (b) What is the change in kinetic energy from point B to point C? (Ans.: -5.88 J) (c) What is the velocity of the box at point C? (Ans.: 22.79 m/s) (d) What is the length of the incline, f? (Ans. 34.88 m) (e) What is the change in gravitational potential energy from point A to point D? (Ans. 17.10 J) www C D A m X B

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