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# Law of conservation of energy

AP.PHYS:
CON‑5.B (EU)
,
CON‑5.B.1 (EK)
,
CON‑5.B.3.1 (LO)
,
CON‑5.B.4 (EK)
,
CON‑5.B.4.1 (LO)
Sal discusses how energy can't be created or destroyed in an isolated system, and works an example of how energy is transformed when a ball falls toward the Earth.

## Want to join the conversation?

• At , why does the kinetic energy turn into thermal energy in the form of heat when touching the ground?
• I believe that the kinetic energy turns into thermal energy due to the friction from the air resistance. Ai resistance produces heat, so some of the kinetic energy turns into heat when touching the ground. Hoope that helps!
• So theoretically, the spring-earth-ball system will go on forever?
• Yes as long as it's just a ball falls perpendicular to a surface with no dissipating forces (like air resistance) to take that energy away.

In the real world, this would be impossible without something putting work to the ball to replace the lost energy.
• At when discussing the ball falling onto the spring, shouldn't the ball still have some gravitational potential energy, since the ball is still at some appreciable height above the ground? The video says that all of the mechanical energy is due to Kinetic energy at the moment of impact.
• What about when you drop something more dense like a bowling ball does the potential energy convert onto kinetic energy?
• The law of conservation of energy holds true regardless of density. So yes, a bowling ball will also have its initial potential energy converted into kinetic energy.
• At , can the energy be also dissipated in form of sound energy?
• Yes that is also possible. Energy is needed for a sound wave to propagate through a medium.
(1 vote)
• What about black holes? The gravitational pull of black holes is so immense that even light cannot escape it. What happens to light inside the black hole? Is it transformed into another type of energy, absorbed as heat, or just... gone?
• Nobody knows. The light is compressed into the black hole's singularity, getting compressed infinitely small.
(1 vote)
• W = F*D.
So with a force of 5N, mass of 1 and acceleration of 5m/s, for one meter I have a work equals to 5J.

The kinetic energy is then half the mass times the velocity square; half the mass is 0.5 which times 5^2(25) is 12.5J.

How did a work of 5 joules equals a kinetic energy of 12.5 joules? Should it not conserve the five joules of energy inserted on it?

Since the first lesson I have this same problem, but now nearing the end I still could not find what I am missing or how to relate kinetic energy to anything else.
(1 vote)
• You're right that with those givens (f=5N, m=1kg, a=5m/s², Δx=1m) you get a work of 5J.

There are kind of two mistakes with your kinetic energy there, though. The first is that it seems you're assuming but not directly stating that the initial velocity (which changes when the work is applied) is zero. Remember, the final kinetic energy will depend on both the initial kinetic energy (found from the initial velocity) and how much the kinetic energy changes (found from the work done).

Second, I think you made a mistake in calculating the final velocity for the final kinetic energy. In this situation, you can use a kinematic equation to solve for the final velocity. We'll say that the initial velocity is actually v0=0m/s and our other relevant givens are a=5m/s² and Δx=1m. We want to find the final velocity (v), so we can use the equation v²=v0²+2aΔx to get v=√(v0²+2aΔx).

Plugging in the given values gets us v=√(0²+2*5*1)=√10m/s. Putting that into the kinetic energy equation, we get Ek=½mv²=½(1)(√10)²=5J. The final kinetic energy is the same as the work done as long as the initial velocity is 0m/s.

To get a more general answer, we can substitute the expression for the final velocity (√(v0²+2aΔx)) into the kinetic energy equation Ek=½mv² to get Ek=½m√(v0²+2aΔx)². This simplifies to get Ek=½m(v0²+2aΔx)
Ek= ½mv0²+½m2aΔx
Ek = ½mv0² + maΔx
Ek = Ek0 + W, which makes sense because it's basically the work-energy theorem.

Does that help a bit?
• Then why doesn't perpetually motion devices work?
(1 vote)
• The entropy in the system will always increase. The longer the machine runs the more energy will be converted into heat until the system can no longer sustain its motion.
• It's been said that the universe is energy is this statement true being that matter is the "stuff"of the universe and all matter has energy? To me it makes sense to say yes but I need a definitive answer
(1 vote)
• You have to consider that just because someone ate tofu doesn't mean they're now tofu, though people believe that matter and energy are the different interactions by the same something. (Take my answer with a grain of salt though. I don't know too much about relativity. I sure hope someone with expertise can help.)
• A closed system is where energy does not go in or out.
An open system is where energy can leave and come in.

Does the law of conservation of energy apply to open systems?

Thanks :D
(1 vote)
• In a sense it does. If you keep work in you equation you can still use a form of it. Like Ug + W = K. In that example the gravitational energy plus the work done on the system equals the final kinetic energy. You would use something like that in an open system