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

by talaroue
Tags: breaks, lands, safe, spring, string
 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.
 P: 303 any ideas?
 Sci Advisor HW Helper PF Gold P: 6,040 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.
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P: 26,148
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 by talaroue 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.
 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?
 Admin P: 21,875 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.