The figure shows a plot of potential energy U versus position x of a 0.300 kg pa

Question

The figure shows a plot of potential energy U versus position x of a 0.300 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 8.00 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?

The figure shows a plot of potential energy U versus position x of a 0.300 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 ½potential hill½ of ½height½ UB = 12 J, with kinetic energy 8.00 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

(a) Total energy of the particle = 12 + 8 = 20 J
Kinetic energy of teh particle when it is at x = 3.5 = 20 - 9 = 11J
Let speed be v. We have 0.5*0.30*v^2 =11

v = 8.56 m/s

(b) At x= 6.5, U=0. All the energy is kinetic. so kE=20j

so 1/2 (0.3)v2=20

v= 11.55 m/s.

(c )turning point at right, say x = 'XR' will be where U(XR) = 20 J

slope of U(x) from x =7 to 8 is (24-0)/21= 24 J/m . We have
U(7) = 0 U(XR) = 20 and (20 - 0)/(XR - 7) = 24or
XR-7 = (20/24) or XR = 7 + (20/24) = 7.83 m

(d) turning point at left, say x = 'XL' will be where U(XL) = 20 J
slope of U(x) from x =1 to 3 is (20-11)/2 = 4.5J/m . We have
U!1) = 20 U(XL) = 20and (20 -20)/(XL - 1) = 5.5 or
XL -1 = (2/5.5) or XL = 1 + (2/5.5) = 7.5/5.5 = 1.364 m

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