The figure shows a plot of potential energy U versus position x of a 0.250 kg pa

Question

The figure shows a plot of potential energy U versus position x of a 0.250 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9.00 J, 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 8.00 J. 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

Total starting energy is 20 J = 12 J + 8 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
20J -9J = 11J.

(1/2)(0.25kg)*v^2 = 11J

v^2 = 2*4*11J/kg = 88J/kg = 88(m/s)^2

v = sqrt(88)(m/s) = 9.38 m/s

b) At x = 6.5, the potential energy is zero, so the full 20J of total energy must equal the kinetic energy.

You can use part (a) to work this out the same way, but with the number on the right starting at 20J, instead of 9J.

c) The slope above x = 6.5 m is 24J/m. To rise to a height where all the 20J 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

20J/(24J/m) = 0.834 m past the 6.5m to

x = 7.334 m

d) At x = 3.5 m and moving downward, the potential energy starts at 9J at x = 3.5 m, and it needs to gain 9J more to convert all the 20 J 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 3.5 m by
(11J/2m)*y = 9J , and
y = 9J/(11J/2m) = ?? m below x = 3.5. You should be able to finish it from this point.

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