Consider the potential energy graph shownThis graph is for a 2 kg object. At t=0 youare located at point D and moving with aspeed of 6 m/s10090 AK(JHBFind the total mechanical energyDCompare the speed at each of thelabeled pointsLocate any turning points. (They arenot necessarily on the labeled points.)F20G10001235678 910 114Position (m)Potential Energy (J)

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Asked Nov 13, 2019
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Consider the potential energy graph shown
This graph is for a 2 kg object. At t=0 you
are located at point D and moving with a
speed of 6 m/s
100
90 A
K
(J
H
B
Find the total mechanical energy
D
Compare the speed at each of the
labeled points
Locate any turning points. (They are
not necessarily on the labeled points.)
F
20
G
10
0
0
1
2
3
5
6
7
8 9
10 11
4
Position (m)
Potential Energy (J)
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Consider the potential energy graph shown This graph is for a 2 kg object. At t=0 you are located at point D and moving with a speed of 6 m/s 100 90 A K (J H B Find the total mechanical energy D Compare the speed at each of the labeled points Locate any turning points. (They are not necessarily on the labeled points.) F 20 G 10 0 0 1 2 3 5 6 7 8 9 10 11 4 Position (m) Potential Energy (J)

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Expert Answer

Step 1

The total mechanical energy at point D for t=0, and be calculated as,

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ED KED+PED 1 mv2PE 2 (2 kg)(6 m/s) +(40J) 1 J (36 kg m2 +(40 J) /s2 1 kg m2/s =76 J

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Step 2

From the above it can be inferred that when the potential energy increases, the kinetic energy decreases. Thus, the speed also decreases.

 

From the graph, the speed at each of the labeled points can be compared as,

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РЕ, — РЕx > РЕ, > РЕ# > РЕ; > РЕ, 3 РЕ, > РЕ, > РЕС > РЕ, > РЕ, Thus КЕ, > КЕ, > КЕ, > КЕ, > КЕ, %3 KЕ, > КЕ, > КЕ, > КЕ, > КЕ, — KE, Thus Ve> VF >Vc> W> V, > V3 >Vң>V>3Vx

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Step 3

Consider any turning point t. At the turning point, the speed will be zero. So, the kinetic energy at this point wi...

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E 3D КЕ, + PE 1 -mу* + PE, Eр 2 то) « РЕ, 1 m(0)+PE, ED 2 Е, %3D РЕ, РЕ, — 76 J

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