- #1
blintaro
- 37
- 1
Hi guys,
I work at the tutoring center of my university, and recently a student came in with a question that has been troubled me ever since.
Say you have a certain mass M a distance Δχ above a massless spring. When asked how much energy this system has, you would say it has only gravitational potential energy related to the mass's distance above your P = 0 line, as the mass is not moving, and the spring itself is at rest in equilibrium.
After you drop the mass, considering the precise moment where it latches inelastically onto the spring, wouldn't the mass now not only have that original potential energy (now part converted to kinetic) but also spring energy related to the distance of the latched mass from the new equilibrium point? So have we somehow "added" energy? (Further: would't the mass accelerate faster than g downward toward the new equilibrium point?)
I work at the tutoring center of my university, and recently a student came in with a question that has been troubled me ever since.
Say you have a certain mass M a distance Δχ above a massless spring. When asked how much energy this system has, you would say it has only gravitational potential energy related to the mass's distance above your P = 0 line, as the mass is not moving, and the spring itself is at rest in equilibrium.
After you drop the mass, considering the precise moment where it latches inelastically onto the spring, wouldn't the mass now not only have that original potential energy (now part converted to kinetic) but also spring energy related to the distance of the latched mass from the new equilibrium point? So have we somehow "added" energy? (Further: would't the mass accelerate faster than g downward toward the new equilibrium point?)