The elastic potential energy stored in a spring is equal in magnitude to the work done by it when moving from its compressed distance, x, to its original uncompresed length, that is, [tex]E_p = \int Fdx[/tex], from the definition of work, where F, from Hookes Law, is [tex] F =kx [/tex]. Performing the calculus, you get the well known equation for the elastic potential energy of the spring, [tex]E_p =1/2kx^2[/tex]. Noting that [tex]F=kx[/tex] at its max compressed point, this is identically equivalent to [tex]E_p = F/2 (x) [/tex], or [tex]E_p = F_{avg} x[/tex].temaire said:So Phanthom, you're saying that [tex]E_p = F_{ave}x[/tex] is an actual formula?
The Spring Problem is a physics concept that involves calculating the displacement, velocity, and acceleration of an object attached to a spring and released from a certain height.
To solve the Spring Problem, you will need to use the formula F = -kx, where F represents the force exerted by the spring, k is the spring constant, and x is the displacement of the object from its equilibrium position.
The average force in the Spring Problem refers to the average amount of force exerted by the spring on the object as it moves from its initial position to its final position.
The spring constant and average force have a direct relationship. This means that as the spring constant increases, the average force exerted by the spring also increases, and vice versa.
The mass of the object attached to the spring does not affect the spring constant or average force. However, it does affect the displacement, velocity, and acceleration of the object as it moves due to the force exerted by the spring.