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

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Homework Help Overview

The problem involves a 1209 kg safe that falls from a height of 1.54 m onto a spring, compressing it by 53 cm. Participants are discussing how to calculate the spring constant using energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply energy conservation equations, specifically relating gravitational potential energy and spring potential energy. There are questions about the correct mass value and the height to be used in calculations.

Discussion Status

Some participants have pointed out errors in the original poster's calculations, particularly regarding the mass and the height used in the energy equations. There is a recognition of the need to consider the total height when calculating gravitational potential energy. Multiple interpretations of the approach are being explored, with suggestions for correcting the setup.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to accurately account for the height of the safe after it compresses the spring. There is also a mention of homework rules that may limit the type of assistance provided.

talaroue
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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?
 
Last edited:
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"
 
Last edited by a moderator:
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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