Problem 4.1
[Difficulty: 2]
Given:
Data on mass and spring
Find:
Maximum spring compression
Solution:
The given data is
M5
l
b
⋅
=
h5
f
t
⋅
=
k2
5
lbf
ft
⋅
=
Apply the First Law of Thermodynamics: for the system consisting of the mass and the spring (the spring has gravitional potential
energy and the spring elastic potential energy)
Total mechanical energy at initial state
E
1
0
=
Note: The datum for zero potential is the top of the uncompressed spring
Total mechanical energy at instant of maximum compression x
E
2
Mg
⋅
x
−
()
⋅
1
2
k
⋅
x
2
⋅
+
=
But
E
1
E
2
=
so
0M
g
⋅
x
−
⋅
1
2
k
⋅
x
2
⋅
+
=
Solving for x
x
2M
⋅
g
⋅
k
=
x25
×
lb
⋅
32.2
×
ft
s
2
⋅
ft
25 lbf
⋅
×
32.2 lb
⋅
ft
⋅
s
2
lbf
⋅
×
=
x
0.401 ft
⋅
=
When just resting on the spring
x
⋅
k
=
x
0.200 ft
=
When dropped from height h:
Total mechanical energy at initial state
E
1
⋅
h
⋅
=
Total mechanical energy at instant of maximum compression x
E
2
⋅
x
−
⋅
1
2
k
⋅
x
2
⋅
+
=
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 Fall '07
 Lear
 Dynamics, Fluid Mechanics, Thermodynamics, Energy, Force, Potential Energy, 0.200 ft

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