The following graph plots the potential energy as a function of x. A 0.4 kg part
ID: 1649391 • Letter: T
Question
The following graph plots the potential energy as a function of x. A 0.4 kg particle is released at point b (x = 2.5 m) with a kinetic energy of K = 4 J, traveling to the right. Assume UA = 9 J, UB = 12 J, UC = 20 J, and UD = 24 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. At what position x will the particle stop and turn around?
d. After turning around, the particle will be traveling to the left. At what position will it stop and turn around again?
Up 0 x (m)Explanation / Answer
Total starting energy is 16 J = 12 J + 4J. 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
16J -9J = 7J.
(1/2)(0.4kg)*v2 = 7J
v2 = 2*2.5*7J/kg = 35J/kg = 35(m/s)^2
v = sqrt(2*2.5*7)(m/s)
v = 6 m/s
b) At x = 6.5, the potential energy is zero, so the full 16J of total energy must equal the kinetic energy.
(1/2)(0.4kg)*v2 = 16J
v2 = 2*2.5*16J/kg = 80J/kg = 80(m/s)^2
v = sqrt( 2*2.5*16)(m/s)
v = 9 m/s
c) The slope above x = 6.5 m is 24J/m. To rise to a height where all the 16J 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
16J/(24J/m) = 0.66m past the 6.5m to
x =6.5+.66= 7.16m
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 16J of total energy into pure potential energy and stop.
The slope is(20J-9J)/(3m-1m)= 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) = 1.63 m below x = 3. You should be able to finish it from this point and turn around again.
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