In conservation of energy, for example throwing a ball up and catching it, how does air resistance (non free fall) impact the conservation of energy? Is it because energy is lost due to work done on the air?
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In conservation of energy, for example throwing a ball up and catching it, how does air resistance (non free fall) impact the conservation of energy? Is it because energy is lost due to work done on the air?
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5 Responses
correct.
the total energy of the ball will be the sum of its remaining mechanical energy plus, the energy lost due to work done against air resistance. so, still the conservation of energy holds good.
Energy is ALWAYS conserved if you count everything. So, any energy lost by the ball in coming back down slower is gained by the heating up of the air as the ball collides with air molecules and speeds them up (causing an increase in the temperature of the air, or an increase in the heat energy of the air.)
Short answer, yes, work is done on the air, too.
When I went to school, “The Conservation of Energy” meant simply that energy can not be created or destroyed.
Therefore to answer the question asked, the effect of air resistance has really NO effect on the “Conservation of Energy”, as I understand it.
IF by conservation of energy U meant instead, the “conservation of USEFUL work/energy” THAT depends on the presence or lack of friction. Friction work is “lost” in the sense that it bleeds useful energy from the system. Such energy (friction) can never again be used to do useful work.
“What is the effect of air resistance on the conservation of energy?”
It doesn’t change the idea of conservation of energy, but it makes calculations using it more difficult.
“Is it because energy is lost due to work done on the air?”
Yes. Air resistance creates lower final energy values than we calculate if we neglect it.
So the ball wouldn’t go as high or fall back as fast as you would predict if you neglected air resistance.
Without air resistance conservation of energy tells us that:
mv^2/2 = mgh
where v is the velocity at which the ball left your hand and h is the maximum height. The mass m cancels and we can write:
v^2/2 = gh, so h = v^2/2g
If we consider air resistance we have to write:
mv^2/2 = mgh + A
Where A is the energy lost to air resistance.
Notice that mass no longer cancels. We can solve for h :
v^2/2g – A/mg = h
this is less than
v^2/2g