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

In summary, when a 1209 kg safe falls 1.54 m and compresses a heavy-duty spring by 53 cm, the spring constant can be found by setting the initial gravitational potential energy equal to the final energy, where the initial and final velocities are both zero. The correct formula to use is k = 2mgh/Δx^2, after correcting for the safe's mass and considering its final height as 2.07 m.
  • #1
talaroue
303
0

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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  • #2
any ideas?
 
  • #3
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.
 
  • #4
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:
 
  • #5
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:
  • #6
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.
 
  • #7
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'.
 
  • #8
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:
  • #9
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.
 

1. How does a safe falling on a spring relate to science?

The scenario of a safe falling on a spring involves the principles of gravity, potential energy, and motion, which are all fundamental concepts in the field of physics.

2. Why does the safe break when it falls on the spring?

The safe breaks because of the force of impact when it hits the spring. The spring compresses and exerts an equal and opposite force on the safe, causing it to break due to the stress and strain on its structure.

3. Is there a specific type of spring that is more likely to break the safe?

The type of spring used does not necessarily determine whether the safe will break. The force of impact and the strength of the safe's structure are the main factors that determine if it will break or not.

4. Can the string prevent the safe from breaking?

The string may help reduce the force of impact on the safe, but it cannot completely prevent it from breaking. The string may also break due to the weight of the safe, making it unable to support it.

5. What other factors can affect the outcome of the safe falling on the spring?

The height from which the safe falls, the material and design of the safe, and the angle at which it hits the spring can all affect the outcome of the safe breaking or not.

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