Vertical Spring


by BuBbLeS01
Tags: spring, vertical
BuBbLeS01
BuBbLeS01 is offline
#1
Oct25-07, 08:47 PM
P: 605
1. The problem statement, all variables and given/known data
A 315.0 g block is dropped onto a vertical spring with a spring constant k = 262.0 N/m. The block becomes attached to the spring, and the spring compresses 0.12 m before momentarily stopping. While the spring is being compressed, what work is done by the block's weight?

What work is done by the spring?

What was the speed of the block just before it hit the spring?


2. Relevant equations
Us = 1/2kx^2


3. The attempt at a solution
Us = 1/2 * 262 *.12^2
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Mindscrape
Mindscrape is offline
#2
Oct25-07, 09:11 PM
P: 1,877
Should be right, what is your question?
BuBbLeS01
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#3
Oct25-07, 09:13 PM
P: 605
That is the work done by the spring right? Because it's saying it's wrong. I am getting an answer of 1.886N*m.

BuBbLeS01
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#4
Oct25-07, 10:07 PM
P: 605

Vertical Spring


I don't understand why that equation is not working.
Mindscrape
Mindscrape is offline
#5
Oct25-07, 10:18 PM
P: 1,877
Are you sure that the units are all what you posted? Does it want you to define your zero potential in a certain way, has it located the origin for you, i.e. might it care about signs?

Yes,

F_s = -kx, so W_s = U_s = integral(-dU) = integral(-kx) = .5*kx^2
BuBbLeS01
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#6
Oct25-07, 10:52 PM
P: 605
oh ok I got it..so for part do you just multiply that by the weight of the block?
Mindscrape
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#7
Oct26-07, 01:56 AM
P: 1,877
You mean how do you find the speed of the block? It screams conservation of energy.
BuBbLeS01
BuBbLeS01 is offline
#8
Oct26-07, 06:13 AM
P: 605
Ug = mgy

So is it just the formula for gravitational energy?
Jealous
Jealous is offline
#9
Oct20-10, 07:42 PM
P: 1
does anyone know the answer to this problem? I have the same issue... One thing to remember they block is being dropped from a height above the spring and then it compresses the spring so I need to find a way to find that height before it comes in contact with the spring


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