The figure shows a plot of potential energy U versus position x of a 0.8 kg part
ID: 1448823 • Letter: T
Question
The figure shows a plot of potential energy U versus position x of a 0.8 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9.0 J, UB = 12.0 J, UC = 19.0 J, and UD = 22.0 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?
t- 0 x (m)Explanation / Answer
In this problem mechanical energy is conserved as the particle moves
A.
Ub + Kb = Ua + Ka
Ka = 12 - 9 + 4 = 7 J
A is the region from x = 3 to x = 4
x = 3.5 comes between A region so speed at x = 3.5 will be
ka = 0.5*m*va^2
va = sqrt(2*Ka/m)
va = sqrt(2*7/0.8) = 4.183 m/sec
B.
U0 + K0 = Ub + Kb
at x = 6.5, U0 = 0
K0 = 12 - 0 + 4 = 16 J
O is the region from x = 6 to x = 7
x = 6.5 comes between O region so speed at x = 6.5 will be
ka = 0.5*m*va^2
va = sqrt(2*Ka/m)
va = sqrt(2*16/0.8) = 6.324 m/sec
C. At the turning point, the speed of the particle is zero. Let the position of the right turning point be Xr.
Between region x = 7 and x = 8 there will be a turning point.
point will be
O (7.0 m, 0 J)
C (Xr, 16.0 J)
because at Xr all kinetic energy from point O will be converted into potential energy.
D (8.0 m, 22.0 J)
slope
(16 - 0)/(Xr - 7) = (22 - 16)/(8 - Xr)
Xr = 85/11 = 7.72 m
D.
At the turning point, the speed of the particle is zero. Let the position of the left turning point be Xl.
Between region x = 1 and x = 3 there will be a turning point.
point will be
A (3.0 m, 9 J)
O (Xl, 16.0 J)
because at Xl all kinetic energy from point A will be converted into potential energy.
C (1.0 m, 19.0 J)
slope
(16 - 19)/(Xl - 1) = (9 - 16)/(3 - Xl)
Xl = 8/5 = 1.6 m
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