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posted by  magzy on 7/3/2009 2:07:02 PM  |  status: Closed  |  Earned Karma: 140

13.25

Course Textbook Chapter Problem Needs by
Classical Mechanics Vector Mechanics for Engineers: Dynamics (8th) by Beer, Johnston, Clausen 13 25P N/A
Question Details:

An 8-kg plunger is released from rest in the position shown and

is stopped by two nested springs; the constant of the outer spring

is k1 = 3 kN/m and the constant of the inner spring is k2 = 10

kN/m. If the maximum deflection of the outer spring is observed

to be 150 mm, determine the height h from which the plunger

was released.
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Guru
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posted by Balu on 7/4/2009 1:21:51 PM  |  status: Live
Asker's Rating: Helpful   
Response Details:
Using Work and energy 
T1+U1→2=T2 (1)
Where T1=0;T2=0
Work
Outer spring:V1-2=33.75J
Inner spring:U1-2=18J
Gravity U1−2=mg(h+0.15)=(8)(9.81)(h+0.15) = 78.48h+ 11.722
Total work  U1−2= −33.75 −18 + 78.48h+11.772= −39.978 + 78.48 h
Substituting into (1)
0−39.978+78.48 h=0
h = 0.5094 m
h = 509 mm
bhargav
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