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Vertical Spring 
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#1
Oct2507, 08:47 PM

P: 602

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 


#2
Oct2507, 09:11 PM

P: 1,874

Should be right, what is your question?



#3
Oct2507, 09:13 PM

P: 602

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.



#4
Oct2507, 10:07 PM

P: 602

Vertical Spring
I don't understand why that equation is not working.



#5
Oct2507, 10:18 PM

P: 1,874

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 


#6
Oct2507, 10:52 PM

P: 602

oh ok I got it..so for part do you just multiply that by the weight of the block?



#7
Oct2607, 01:56 AM

P: 1,874

You mean how do you find the speed of the block? It screams conservation of energy.



#8
Oct2607, 06:13 AM

P: 602

Ug = mgy
So is it just the formula for gravitational energy? 


#9
Oct2010, 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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