Question

MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSİON 4 BACK Your answer is partially correct. Try again. The figure shows a plot of potential energy U versus position x of a what is the speed of the particle at (a)x = 3.50 m and (b)x-6.50 m? what is the position of the turning point on (c) the right can only along an x axis under the influence of a conservative force. The graph has these values: UA 9.00 3, Uc 20.0 J and Up 24.0 J. The particle is released at the point where U forms a "potential hill" of "height Ug 12.0 J, with kinetic energy 8.00 1 UB Units ['m/s

Answer 1

(A) Initial mechanical or total energy = 12 + 8 = 20 J

At point x = 3.5, potential energy is U_A which is 9 J

therefore, kinetic energy at x = 3.5

K.E = 20 - 9 = 11 J

v = sqrt (2K.E / m )

v = sqrt (2*11 / 0.250)

v = 9.3808 m/s

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(B) For x = 6.5, potential energy is 0

so,

K.E = 20 J

v = sqrt (2*20 / 0.250)

v = 12.65 m/s

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(C) 20 = 24 (x - 7)

x = 7.8333 m

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(D) 20 = (9-20 / 2 ) x - 1 + 20

20 = -5.5 (x - 1) + 20

20 = -5.5x + 5.5 + 20

x = 1 m