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Finding spring compression

by emilykorth
Tags: energy, kinetic, potential, spring
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emilykorth
#1
Oct13-08, 12:09 PM
P: 13
1. The problem statement, all variables and given/known data

A block of mass m = 2.0 kg is dropped from height h = 55 cm onto a spring of spring constant k = 1960 N/m (Fig. 8-36). Find the maximum distance the spring is compressed.


2. Relevant equations

PEspring=1/2 kx2 F=-kx PE gravity=mgh KE gravity=1/2mv2

3. The attempt at a solution
I started by setting PE gravity equal to PE spring so that mgh=1/2kx2.
so then I had (2)(9.8)(.55+x)=(1/2)(1960)((.55+x)^2)
I then solved for x and got x=.55 but this was wrong-why?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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gabbagabbahey
#2
Oct13-08, 12:20 PM
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The maximum spring compression occurs when the spring has absorbed all of the block's mechanical energy, not just it's potential energy.
emilykorth
#3
Oct13-08, 12:24 PM
P: 13
so how can I vary the equation to make it right?
mgh+1/2mv^2=1/2kx^2?????

gabbagabbahey
#4
Oct13-08, 12:33 PM
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Finding spring compression

The initial gravitational potential energy is certainly mgh, but should it have a gravitational potential of -mgx when the spring is compressed a distance x?

Try looking at the total mechanical energy of the whole system, KE_block+PE_gravity+PE_spring...what will be the kinetic energy of the block when the spring reaches maximum compression? What will be it's gravitational potential energy? What was the initial total energy of the block?
emilykorth
#5
Oct13-08, 12:52 PM
P: 13
When the spring reaches maximum compression there would be no kinetic energy since the block will eventually have no velocity and will stop. Would the KE of the spring be Wspring=-1/2kx^2???
gabbagabbahey
#6
Oct13-08, 12:55 PM
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the KE of the spring is negligible since it is assumed that the spring's mass is very small, so you can ignore it (KE=mv^2/2 remember).

The KE of the block will indeed be zero at the maximum spring compression, what will it's gravitational potential be?
emilykorth
#7
Oct13-08, 01:06 PM
P: 13
mgh?
emilykorth
#8
Oct13-08, 01:08 PM
P: 13
gravitational potential of the spring-would it be 1/2kx^2?
gabbagabbahey
#9
Oct13-08, 01:09 PM
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mgh is the potential energy of the block before it is dropped

when the spring is compressed a distance x, how high is the block (assuming that the height of the uncompressed spring is zero)?
emilykorth
#10
Oct13-08, 01:12 PM
P: 13
it would be -x which is what im trying to find
gabbagabbahey
#11
Oct13-08, 01:20 PM
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Right, so the gravitational potential energy of the block at maximum compression should be -mgx correct? So what is the total energy of the block and spring at maximum compression?
emilykorth
#12
Oct13-08, 01:34 PM
P: 13
yes but wouldnt the gravitational PE of the block involve the height that is given since the block is traveling not only the height but the distance that the spring is compressed???? so wouldnt that be mg(h+x)???I have no idea what the potential energy of the block and spring would be when compressed-Im so confused!!!
gabbagabbahey
#13
Oct13-08, 01:52 PM
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The initial total energy of the system is mgh (since the spring is uncompressed and the block has no initial kinetic enrgy)

The final total energy is the sum of the gravitational potential of the block at it's new height (-x) and the potential of the compressed spring (1/2 kx^2)

Since total mechanical energy is conserved, the initial total energy will equal the final total energy; hence

[itex]mgh=-mgx+\frac{1}{2}kx^2[/itex]

Do you understand?
emilykorth
#14
Oct13-08, 02:23 PM
P: 13
ok yes i think so.....so when a spring is compressed it has just PE
thank you
emilykorth
#15
Oct13-08, 02:24 PM
P: 13
so is h just the h given or would it be h+x?
gabbagabbahey
#16
Oct13-08, 03:01 PM
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h is just the h given.


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