Calculating Spring Constant - Force or Energy?

AI Thread Summary
The discussion centers on two methods for calculating the spring constant (k) when a mass m stretches a spring by x meters. The first method equates the spring force (kx) to the gravitational force (mg), yielding k = mg/x. The second method involves using energy principles, equating the gravitational potential energy lost (mgx) to the energy gained by the spring (1/2 kx^2) and the kinetic energy of the mass. A discrepancy arises when using the energy method, resulting in a k value that is half of the one calculated using force. The correct approach for calculating k in this scenario is to use the force method, while the energy method is more suited for dynamic problems involving motion.
Cintdrix
Messages
4
Reaction score
0
I've been being confused lately as to the 2 methods. The example I'm thinking of is when a weight of mass m is hung on a spring and it stretches x meters.

First of all, I know you can equate the spring force (kx) to the force of gravity (mg), to get
k = mg/x

But is it also possible to say that the gravitational potential energy lost (mgx) is equal to the energy gained by the spring (1/2 kx^2)? When I do this, I get a different k which is half the original k and probably wrong. How can you calculate K for this problem using energy?
 
Physics news on Phys.org
The first method F = mg = kx is what you would use to find the spring constant k. The second equation is used for the dynamic problem "How fast is the mass moving at distance x?". To solve, equate the potential energy "lost" by the downward motion to the potential energy of the spring AND the kinetic energy of the object: E = mgx = (1/2 kx^2) + (1/2 mv^2)
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top