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Maximum spring compression problem

by whdahl
Tags: compression, maximum, spring
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whdahl
#1
Dec3-12, 06:55 PM
P: 4
1. The problem statement, all variables and given/known data

A 7.0kg box moving at 6.0m/s on a horizontal, frictionless surface runs into a light spring of force constant 50N/cm .
What is the maximum compression of the spring?


2. Relevant equations
K=1/2mv^2


3. The attempt at a solution
K=1/2(7.0)(6)^2 = 126N

126N/(50N/cm) = 2.52cm

That is apparently incorrect.
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Greg-ulate
#2
Dec3-12, 07:39 PM
P: 72
Quote Quote by whdahl View Post
K=1/2(7.0)(6)^2 = 126N
K=1/2(7.0)(6)^2 = 126 Joules not Newtons
WillemBouwer
#3
Dec4-12, 01:10 AM
P: 82
2. Relevant equations
K=1/2mv^2

What type of equation is this? Moment, energy, work? This will refer to the equation used fr the spring. Your equation you use for the spring part is incorrect that is why your displacement will be inccorect...

whdahl
#4
Dec4-12, 09:33 AM
P: 4
Maximum spring compression problem

It is a work problem
lep11
#5
Dec4-12, 12:54 PM
P: 137
Once again, think it in terms of energy. What happens to the kinetic energy of the box if energy is conserved?
WillemBouwer
#6
Dec4-12, 11:48 PM
P: 82
Okay so what amount of work(kinetic energy) can be done by the box? Use your equation and do exactly the same, just get your unit correct for work/energy.
What do you think happens to all this energy once the box comes to a momentarily stop? Energy cannot get created or destroyed it can only be tranfered from one form to another... So if there is no kinetic energy in the box, and the system is at a standstill, where is all the potential energy stored?
whdahl
#7
Dec5-12, 01:05 PM
P: 4
The kinetic energy from the box is then stored as spring potential energy P=1/2kx^2. I think I see my problem. I have the final equation:

126J=126Nm=1/2(50N/cm)x^2

25200Ncm=50(N/cm)x^2
x^2=504cm
x=22.45cm

Thank you guys, you helped a lot!
WillemBouwer
#8
Dec5-12, 11:36 PM
P: 82
pleasure, great job by the way...


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