extended springs energy and force?

Question

I am having a small problem: If I have a mass m attached to a massless spring with spring constant k and it extends a distance x then at equilibrium

Force down (mg) = force up (kx) BUT when considering energy

change in potential energy (mgx) = energy stored in spring (1/2kx^2)

These two equations are inconsistent - what am I doing wrong?

Dr. Zorro · Accepted Answer

The force balance analysis is ok. The inconsistency is in you energy balance because you are forgetting that if you let your mass stretch the spring, it will have kinetic energy when it reaches the equilibrium point. If you are supporting it to slowly go to equilibrium, you are putting in work yourself.

? · Answer

The work done is defined as W=Fx d
         In this case force is not constant but depend on x,so we have to add work done by moving small distances at a time till we reach the given distance.This is same as integration so the work done               W=Integral(0----->x)F(x)dx
              or         =.......................kxdx=1.2kx^2
             Now about mgx,Since g is almost constant for small distances so PE=mgx .At equilibrium mg=ka if a is the maximum stretch of the spring,so both values will be same.
IVAN

? · Answer

When a spring held vertically and a mass is attached to it is released it comes down to a position where extension is mg/k but it has gained some velocity hence it moves down to extreme position 2mg/k.

falmer · Answer

L=4 cm=0.04m F=2000N paintings=L*F/2=40J you take advantage of F/2 because you decide on the common fee of the rigidity, and the common fee of a linearly turning out to be variable is the maths regularly occurring of two of its values: enable's say 0 N even as the spring isn't prolonged and F=2000N even as that is prolonged => F(regularly occurring)=(F+0)/2=F/2.

extended springs energy and force?

4 Answers