The figure shows a plot of potential energy U versus position x of a 0.230 kg pa

Question

The figure shows a plot of potential energy U versus position x of a 0.230 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9 J, UC = 20 J and UD = 24 J. The particle is released at the point where U forms a “potential hill” of “height” UB = 12 J, with kinetic energy 4.50 J. What is the speed of the particle at (a)x = 3.5 m and (b)x = 6.5 m? What is the position of the turning point on (c) the right side and (d) the left side?

Explanation / Answer

total starting KE = 12+4.5 = 16.5 Joules

part A: at x = 3.5 m

PE at A = 9 J

hence KE = 0.5 mv^2 = 16.5 -9 = 7.5

velocity V = sqrt(2* 7.5/0.23)   

V = 8.07 m/s

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part B:

at x= 6.5

PE = 0

KE = 0.5 mv^2 = 16.5 J

v^2 = 2*16.5/(0.23)

V = 11.97 m/s

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turing point at right, x = xr

U(xr) = 16.5 J

slope from 7 to 8 is 24 J

slope = (16.5-0)/(xr-6.5) = 24

xr = 7.185 m

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turinng point at left x = xL

again slope (20-16.5)/(xL-1) = 4.5

x = 1.77 m

so particle move from 1.77 to 7.185 m

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