Elastic potential energy and work equations Elastic potential energy is $\frac{1}{2} k x^2$ and work is $F \cdot d$. Why these numbers do not evaluate to the same value in a problem?
The change in potential energy is the work done on a spring - $W = \Delta U$. However, every time I do an example I always get that the work is double the elastic potential energy. What am I missing?
If it takes $2 \text{ N}$ of force to displace a spring by $0.2 \text{ m}$ with a spring constant of $10 \text{ N/m}$ then the work is $W_e = 2 \text{ N} \cdot 0.2 \text{ m} = 0.4 \text{ J}$. However, the elastic potential energy stored in the spring is $U_e = \frac{1}{2} 10 \text{ N/m} \cdot (0.2 \text{ m})^2 = 0.2 \text{ J}$.
 A: 
so if potential energy in a spring is 1/2kx^2 and work f*d. Why do
these numbers not come out to the same thing in a problem?

You have to start out with the general definition of work, which is not simply force times displacement, but is
$$W=\int\vec F.d\vec x$$
It only equals $Fx$ if the force it constant and can come out of the integral. But the force exerted by the spring is not constant, if varies linearly with displacement. So, for the spring, since the force is in the same direction as displacement and since $F=kx$
$$W=\int (kx)dx=\frac{1}{2}kx^2$$
Hope this helps.
A: You are missing that the force changes as the spring elongates. It is 0 at the equilibrium position and it is only equal to the final force at the final extension. The factor of 1/2 accounts for the variation in the force, essentially giving you the average force.
A: If the force is $f(x)=-kx$, then the potential energy is $E(x)=\int _{0}^{x} ky dy = k\frac{x^{2}}{2}$ When you use the formula with $\vec{F} \cdot \vec{dl}$ you need to integrate, you cannot just plug in the numbers.
