The figure shows a plot of potential energy U versus position x of a 0.310 kg pa

Question

The figure shows a plot of potential energy U versus position x of a 0.310 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 4.00 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

According to the given problem,

(a) Total energy of the particle = 12 + 4 = 16 J
Kinetic energy of the particle when it is at x = 3.5 = 16 - 9 = 7 J
Let speed be v. We have 0.5*0.310*v2 = 7 or v = sqrt[14/0.310] = 6.72 m/s

(b) speed(at x = 6.5) = sqrt[{2*(16-0)}/0.310] = 10.16 m/s

(c )turning point at right, say x = 'XR' will be where U(XR) = 16 J
slope of U(x) from x =7 to 8 is (24-0)/21= 24 J/m . We have
U(7) = 0 U(XR) = 16 and (16 - 0)/(XR - 7) = 24or
XR-7 = (16/24) or XR = 7 + (16/24) = 7.667 m

(d) turning point at left, say x = 'XL' will be where U(XL) = 16 J
slope of U(x) from x =1 to 3 is (20-9)/2 = 5.5J/m . We have
U!1) = 20 U(XL) = 16 and (20 -16)/(XL - 1) = 5.5 or
XL -1 = (4/5.5) or XL = 1 + (4/5.5) = 1.73 m
So the particle will move from 1.73 m to 7.667 m

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