Understanding Energy and Work Done

AI Thread Summary
Energy is defined as the capacity to do work, and when a rock falls, its energy can be transferred upon impact, potentially producing heat and sound. In a perfectly inelastic collision, momentum is conserved while kinetic energy is not, leading to energy dissipation as heat and sound. The discussion also highlights that work done against friction requires movement; if the pillar does not move, no work is done against friction. Calculating the effect of acceleration on useful work involves applying Newton's second law and adjusting for the necessary work to achieve optimal results. Overall, understanding energy transfer and work done is crucial in analyzing physical interactions.
sgstudent
Messages
726
Reaction score
3
Energy is the capacity to do work. So if I have energy I can do work. But if a rock has 100J from falling down a building to push a pillar into the ground, it might not have enough energy to have any work done to push it in. So what happens to that 100Js? Ignoring air resistance. Since energy can neither be destroyed or created so I'm unsure where energy that isn't used has gone. I'm theorizing that heat and sound is produced from the impact and that's the energy dissipated but I'm not completely sure about it. I'm unsure if the friction of the pillar constitutes into this. But then again work done against friction also requires a distance which is 0 when the pillar doesn't move. Thanks for the help! :)
 
Physics news on Phys.org
yup, you are right. upon impact(assume rock+pillar stick together ie. perfectly inelastic collision, momentum conserved, kinetic energy not conserved, lost as heat and sound). also, work done against friction between the sides of the pillar against the ground(as your pillar gains momentum and hence velocity).
 
how would you calculate the effect of acceleration on the precentage of useful work done?
 
You would use Newton's second law. The same rock from before is traveling down with a magnitude and direction of mass * acceleration.
sumF=ma,
W=FdcosT
W=dmacosT, substituting.
for some arbitrary acceleration, the work done is given by the equation.. If you want to use percentages, divide this number by the necessary work to achieve an optimal result, and multiply by 100. This method will however lead to some level of uncertainty constrained by the assumption that some optimal amount of work is the minimum amount of work needed in a system.
 
jester1989 said:
yup, you are right. upon impact(assume rock+pillar stick together ie. perfectly inelastic collision, momentum conserved, kinetic energy not conserved, lost as heat and sound). also, work done against friction between the sides of the pillar against the ground(as your pillar gains momentum and hence velocity).
But what if the pillar doesn't move. Then won't there be zero work done again friction?
 
未命名.GIF
 
Last edited:
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