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A safe held by a string breaks and lands on a spring 
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#1
May2609, 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 heavyduty 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. 


#2
May2609, 06:28 PM

P: 303

any ideas?



#3
May2609, 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.



#4
May2609, 06:34 PM

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Thanks
P: 26,157

A safe held by a string breaks and lands on a spring
Hi talaroue!
(try using the X^{2} and X_{2} tags just above the Reply box ) 


#5
May2609, 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? 


#6
May2609, 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.



#7
May2609, 06:46 PM

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#8
May2609, 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 


#9
May2609, 06:50 PM

P: 303




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