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Physics, 16.03.2020 18:41 dodieoddle
An object of mass m is lowered at constant velocity at the end of a string of negligible mass. As it is lowered a vertical distance h, its gravitational potential energy changes by ∆Ug = −m g h. However, its kinetic energy remains constant, so that if we define E = K + Ug, we find ∆E = −m g h. Why isn’t the total energy E conserved? 1. Because the universe is accelerating in its expansion, the object is actually at rest and not descending ... the earth moves away as fast as it moves "down." 2. An external force is doing work on the system. 3. In reality, all objects are massless, so that m = 0 and ∆E = 0. 4. The acceleration of the system is zero. 5. The net force on the system is not zero. 6. Ug is defined incorrectly as if gravity were a constant force. 7. The total energy is indeed conserved, since ∆E = ∆Ug. 8. E is useless in real-world examples like this.
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An object of mass m is lowered at constant velocity at the end of a string of negligible mass. As it...
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