The figure below shows a plot of potential energy (U = PE) versus position (x) o
ID: 1345207 • Letter: T
Question
The figure below shows a plot of potential energy (U = PE) versus position (x) of a 0.200 kg particle that can travel only along an x-axis under the influence of a conservative force. The graph has values UA = 9.00 J, UC = 20.00 J, and UD = 24.00 J. The particle is released at the point where U forms a "potential hill" of "height" UB = 12.00 J, with a velocity of 5.07 m/s in the negative x-direction.
(1) What will be the speed of the particle at x = 3.5 m?
(2) What is the force on the particle in the region between x = 1.0 m and x = 3.0 m?
Explanation / Answer
a)
The energy of the particle is at point B is =12J+(1/2)mv2 =12+0.5*0.200*(5.07)2 =14.570J
From the graph U =UA (at x =3.5m)
=9J
Therefore the kinetic energy isKE = Etot-U =14.570-9 =5.570J
(1/2)mv2 =5.570J
v2 =2*5.570/(0.200) =55.7049 ====>v =7.463m/s
B)
The particle turns around when kinetic energy becomes zero, i.e at the point where the potential energy is equal to the total mechanical energy
PE =Etot =14.570J (between the x =1m to x =3m)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.