The figure here shows a plot of potential energy U versus position x of a 0.874
ID: 2260858 • Letter: T
Question
The figure here shows a plot of potential energy U versus position x of a 0.874 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15 J, UB = 35 J and UC = 45 J. The particle is released at x = 4.5 m with an initial speed of 7.0 m/s, headed in the negative xdirection. (a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of x = 4.0 m? Suppose, instead, the particle is headed in the positive x direction when it is released atx = 4.5 m at speed 7.0 m/s. (d) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? What are the (e)magnitude and (f) direction of the force on the particle as it begins to move to the right of x = 5.0 m?
Explanation / Answer
a) Potential energy required = 35 -15 = 20J
kinetic energy = 0.5mv^2 = 21.41J so it is more than 20J so it can go to x=1. with speed = sqrt(2*1.41/0.874) = 1.798m/s
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