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(d) What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m? magnitude direction N ---Select--- O

(d) What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m? magnitude direction N ---Select--- O

College Physics
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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Need help with part D.

The figure below shows a plot of potential energy U versus position x of a 0.96 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are UA = 10.0 J, UB 25.0 J, and Uc = 35.0 J. = (0) Ucr UB UA 4 x (m) The particle is released at x = 4.5 m with an initial speed of 8.5 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? 6.4 m/s ◆ 6 (b) What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m? 7.5 N magnitude direction magnitude direction +X Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 8.5 m/s. (c) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? 4.5 m/s ◊ (d) What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m? N ---Select--- î
Transcribed Image Text:The figure below shows a plot of potential energy U versus position x of a 0.96 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are UA = 10.0 J, UB 25.0 J, and Uc = 35.0 J. = (0) Ucr UB UA 4 x (m) The particle is released at x = 4.5 m with an initial speed of 8.5 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? 6.4 m/s ◆ 6 (b) What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m? 7.5 N magnitude direction magnitude direction +X Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 8.5 m/s. (c) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? 4.5 m/s ◊ (d) What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m? N ---Select--- î
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