Spring Speed Problem: Solve 4kg Block, 3500N/m Stiffness

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The discussion revolves around solving a physics problem involving a spring launching a 4 kg block. The spring has a stiffness of 3500 N/m and is compressed 0.2 m, with the block released from rest. The key equation used is the conservation of energy, incorporating changes in spring potential energy, gravitational potential energy, and kinetic energy. A common mistake identified was miscalculating the change in spring potential energy and the signs in the gravitational potential energy equation. After correcting these errors, the participant successfully found the speed of the block to be 3.09 m/s at a height of 1.3 m.
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Homework Statement


A spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight up in the classroom. The spring is initially
compressed 0.2 m, and the block is initially at rest when it is released. When the block is 1.3 m above its starting
position, what is its speed?


Homework Equations


System: block, spring and earth.
delta_Uspring + delta_Ugravitation + delta_K=0


The Attempt at a Solution


everything I've tried seems to be wrong, the answer answer given to us however is 3.09 m/s.
 
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timmy8, welcome to PF!

Your equation is correct, so please show your work so we may see where you may have gone wrong.
 
1/2(3500)(1.3^2)-1/2(3500)(0.2^2)+4(-9.8)(1.3-0.2)+1/2(4)Vf^2-1/2(4)(0^2)=0

from there i solved for vf^2;

and vf^2=-1422.19

i am getting Vf^2 to be a negative number and i would have to solve the square root of that to get Vf.
what am i doing wrong?
 
timmy8 said:
1/2(3500)(1.3^2)-1/2(3500)(0.2^2)+4(-9.8)(1.3-0.2)+1/2(4)Vf^2-1/2(4)(0^2)=0

from there i solved for vf^2;

and vf^2=-1422.19

i am getting Vf^2 to be a negative number and i would have to solve the square root of that to get Vf.
what am i doing wrong?
You have the change in the spring potential energy wrong. What is the PE of the spring when the block rises 1.3 m above its start point? The block is not attached to the spring. Also, you seem to have slipped up on a plus /minus sign for the change in gravitational PE. Also, the problem asks for the speed at 1.3 m above the initial start position of the block (when the spring is initially compressed, I think).
 
so in the spring potential should Uf be equal to zero? and for the change in gravitational PE isn't the formula m*g*delta_h? I am so lost.
 
timmy8 said:
so in the spring potential should Uf be equal to zero?
yes
and for the change in gravitational PE isn't the formula m*g*delta_h? I am so lost.
Yes, the formula is change in Gravitational PE is (mgh_final) - (mgh_initial). final h is 1.3 and initial h is zero, so the change is positive (it has gained grav. PE).
 
i thought g would be -9.81?
 
now i have:
-springPE +mgh_final+1/2(m)v_final^2=0

This gave me the correct answer.

thank you for all your help PhantomJay
 
timmy8 said:
now i have:
-springPE +mgh_final+1/2(m)v_final^2=0

This gave me the correct answer.

thank you for all your help PhantomJay
You're welcome. The plus and minus sign can be a killer in Physics, and it is easy to get confused. Sometimes you just have to reason it out.
 
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