(8c8p40) The figure shows a plot of potential energy U versus position x of a 1.
ID: 1443077 • Letter: #
Question
(8c8p40) The figure shows a plot of potential energy U versus position x of a 1.7 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 16.0 J, UB = 19.0 J, UC = 26.0 J, and UD = 31.0 J. The particle is released at the point where U forms a "potential hill" of "height" UB, with kinetic energy 4.0 J. a)What is the speed of the particle at x = 3.5 m?
b)What is the speed of the particle at x = 6.5 m?
c)What is the position of the turning point on the right side?
d)What is the position of the turning point on the left side?with exp if possible
Explanation / Answer
Total starting energy is 23 J = 19 J + 4 J. This total must remain constant since the potential-energy field is conservative.
a) At x = 3.5 m, the potential energy is 16 J, that of the A level, so the kinetic energy must be
23J - 16J = 7J.
(1/2)(1.7kg)*v^2 = 7J
v^2 = 8.24 J/kg
v = 2.87 m/s
b) At x = 6.5, the potential energy is zero, so the full 23 J of total energy must equal the kinetic energy.
(1/2)(1.7kg)*v^2 = 23J
v^2 = 27.06 J/kg
v = 5.2 m/s
c) The slope above x = 6.5 m is 31J/m. To rise to a height where all the 23J 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
23J/(31J/m) = 0.74 m past the 6.5 m to x = 7.24 m
d) At x = 3m and moving downward, the potential energy starts at 16J at x = 3m, and it needs to gain 16J more to convert all the 23J 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
(10J/2m)*y = 16 J , and
y = 16J/(10J/2m) = 3.2 m below x = 3.
Hope it is correct. I tried my best as there is no figure.
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