A safe held by a string breaks and lands on a spring


by talaroue
Tags: breaks, lands, safe, spring, string
talaroue
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#1
May26-09, 05:39 PM
P: 303
1. The problem statement, all variables and given/known data
A 1209 kg safe is 1.54 m above a heavy-duty spring when the rope holding the safe breaks. Thr safe hits the spring and compress it 53 cm. What is he spring constant of the spring?


2. Relevant equations
K=mV^2/2
U=mgh
Ki+Ui=Kf+Uf


3. The attempt at a solution

I thought I understood momentuem and energy but not anymore. Here is what I did....

Ui=mgh= 2*9.8*1.54
Kf=mV^2/2

Then solved for V and got 5.494 m/s

Then used that velocity for the spring constant by using......

U=k(delta x)^2/2
K=mV^2/2

then solved for k and got 129912 N/m...... but that is coming up wrong what am i doing wrong.
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talaroue
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#2
May26-09, 06:28 PM
P: 303
any ideas?
PhanthomJay
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#3
May26-09, 06:33 PM
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Well for one thing you are using m = 2 kg when it is given that m = 1209 kg. Then you neglect the gravitational potential energy in the 2nd set of equations. It is better to do this all in one step: Initial energy = Final energy, where v_initial = v_final = 0.

tiny-tim
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May26-09, 06:34 PM
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A safe held by a string breaks and lands on a spring


Hi talaroue!

(try using the X2 and X2 tags just above the Reply box )
Quote Quote by talaroue View Post
A 1209 kg safe is 1.54 m above a heavy-duty spring when the rope holding the safe breaks. Thr safe hits the spring and compress it 53 cm.

Ui=mgh= 2*9.8*1.54
No, the safe stops moving 53cm lower.
talaroue
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#5
May26-09, 06:36 PM
P: 303
JAY:

so your saying that i should just use U=mgh and U=k(delta x)^2/2....solve for k and get....

k=2mgh/delta x^2

TINY TIM:

so I didn't go about the problem wrong I just have to add the .53 m to the hieght?
Astronuc
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May26-09, 06:42 PM
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The safe starts 1.54 m (154 cm) above the spring and then travels another 0.53 m after contact. That must be considered with respect to gravitational potential energy.
PhanthomJay
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#7
May26-09, 06:46 PM
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Quote Quote by talaroue View Post
JAMES:

so your saying that i should just use U=mgh and U=k(delta x)^2/2....solve for k and get....

k=2mgh/delta x^2
Yes, but 'h' is as noted by others, and don't forget to correct your value of 'm'.
talaroue
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#8
May26-09, 06:47 PM
P: 303
ahhh I see at least I had the right idea just need to be more careful. If you guys don't mind I am having problems with another problem as well.... it might be a same problem i am going to go back and look at it but I made a thread about it as well....

Second problem I have issues with
talaroue
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#9
May26-09, 06:50 PM
P: 303
Quote Quote by PhanthomJay View Post
Yes, but 'h' is as noted by others, and don't forget to correct your value of 'm'.
I wasn't using 2 as my m, that was just noting the second equation is U=k(delta x^2)/2 I didn't combined them when i posted that, i see what you are saying though.


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