The figure shows a plot of potential energy U versus position x of a 0.290 kg pa

Question

The figure shows a plot of potential energy U versus position x of a 0.290 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 6.00 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?

I know that the answer to (a) is 7.87m/s, (b) is 11.1m/s, and (d) is 1.36m. I do not understand why I could not get the answer to (c).

Explanation / Answer

Total starting energy is 12 J + 6 J = 18 J. This total must remain constant since the potential-energy field is conservative.

a) At x = 3.5 m, the potential energy is 9 J, that of the A level, so the kinetic energy must be

KE = 18J -9J = 9J.

(1/2)*(0.29kg)*v^2 = 9J

v^2 = 2*9/(0.29)= 62.07 (m/s)^2

v = 7.87 m/s

b) At x = 6.5, the potential energy is zero, so the full 18J of total energy must equal the kinetic energy.

so KE = 18 J

(1/2)*(0.29kg)*v^2 = 18J

v^2 = 2*18/(0.29)= 124.13 (m/s)^2

v = 11.14 m/s

c) The slope above x = 6.5 m is 24J/m. To rise to a height where all the 18J of total energy is converted into potential energy and the mass stops (kinetic energy is zero, so v must be zero), the mass must rise

18J/(24J/m) = 0.75m past the 6.5m to

x = 6.5 + 0.75 = 7.25 m

the position of the turning point on the right side = 7.25 m

d) At x = 3m and moving downward, the potential energy starts at 9J at x = 3m, and it needs to gain 9J more to convert all the 18J of total energy into pure potential energy and stop.

The slope is 11J/2m for the rise moving downward from x = 3m. So the mass must go down from 3m by

(11J/2m)*y = 9J , and

y = 9J/(11J/2m) = 1.64 m below x = 3.

so the position of the turning point on the left side = 3 - y = 3 - 1.64 = 1.36

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