Two equal-length springs are \"nested\" together in order to form a shock absorb
ID: 1719005 • Letter: T
Question
Two equal-length springs are "nested" together in order to form a shock absorber.
(Figure 1)
Part A
If it is designed to arrest the motion of a 3-kg mass that is dropped s = 0.5 m above the top of the springs from an at-rest position, and the maximum compression of the springs is to be 0.2 m, determine the required stiffness of the inner spring, kB, if the outer spring has a stiffness kA = 370 N/m .
Express your answer to three significant figures and include the appropriate units.
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Figure 1 of 1
Two equal-length springs are "nested" together in order to form a shock absorber.
(Figure 1)
Part A
If it is designed to arrest the motion of a 3-kg mass that is dropped s = 0.5 m above the top of the springs from an at-rest position, and the maximum compression of the springs is to be 0.2 m, determine the required stiffness of the inner spring, kB, if the outer spring has a stiffness kA = 370 N/m .
Express your answer to three significant figures and include the appropriate units.
kB =SubmitMy AnswersGive Up
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Figure 1 of 1
Explanation / Answer
Given:
Ka= 370 N/m
M= 3 Kg
s= 0.5 m
g = 9.81 m/s2
s1= 0.2 m
Assume that we compress the Combo spring sufficiently, so that Spring B is compressed exactly 1.0 meter. Now we can find the distance "x" that Spring A is compressed:
Fa = Fb
(Ka)(x) = (Kb)(1 meter)
370 N/m(x) = (Kb) N/m
x = kb/370 meters
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Now we can find Kc for the combo-spring:
Force exerted by combo spring = kbN/m(1m) + 370N/m(kb/370m) = kb+ kb = 2kb
Compression distance of combo-spring = 1+ kb/370 m
Therefore Kc = 2kb/(1+kb/370)
Potential energy of the block is converted into spring energy
mgs = -kc (s1)
3*9.81*0.5 = 2kb/(1+kb/370) *0.2
on solving Kb= 40.84 N/m
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