The figure shows a plot of potential energy U versus position x of a 0.260 kg pa

Question

The figure shows a plot of potential energy U versus position x of a 0.260 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: Ug = 9.00 ], Uc = 20.0 J and UD = 24.0 J. The particle is released at the point where U forms a "potential hill" of "height" UB-12.0 J, with kinetic energy 3.50 . 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 side and (d) the left side?

Explanation / Answer

from the given data

mass of particel, m = 0.26 kg

Ua = 9 J

Ub = 12 J

Uc = 20 J

Ud = 24 J

initial release from point B, at KE = 3.5 J

so total inital energy E = KE + Ub = 3.5 + 12 = 15.5 J

a. at x = 3.5 m

Ua = 9 J

let speed be u

then from conservatioon of energy

0.5mu^2 + 9 = 15.5

0.5*0.26*u^2 = 15.5 - 9 = 6.5

hence u = 7.071 m/s

b. at x = 6.5 m, just like previous part

0.5*0.26*u^2 + 0 = 15.5

u = 10.9192 m/s

c. position of turning point in the right side be x

then, Ux = 15.5 J

(Ux - 0)/(x - 7) = (Ud - 0)/(8 - 7)

x = 7.6458 J

d. position of turning poiont on the left side = x

then

Ux = 15.5 J

(Ux - Ua)/(x - 3) = (Uc - Ua)/(1 - 3)

(15.5 - 9)/(x - 3) = -(20 - 9)/2

x = 1.818 m

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