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

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A 1209 kg safe falls 1.54 m onto a spring after its rope breaks, compressing the spring by 53 cm. The initial potential energy (Ui) is calculated using Ui = mgh, while the final energy involves the spring's potential energy (U) given by U = k(delta x)^2/2. The main error identified in the calculations is the incorrect mass used and the need to account for the total height when calculating gravitational potential energy. The correct formula for the spring constant (k) is derived as k = 2mgh/delta x^2, emphasizing the importance of using the correct values for mass and height. Understanding these concepts is crucial for solving similar physics problems effectively.
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Homework Statement


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?


Homework Equations


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


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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any ideas?
 
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.
 
Hi talaroue! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)
talaroue said:
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. :wink:
 
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?
 
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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.
 
talaroue said:
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'.
 
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...

https://www.physicsforums.com/showthread.php?t=316282"
 
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PhanthomJay said:
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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