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.

Mechanical would have been conserved if only the force of gravity (the weight of the object does work on the system). The tension force does work also on the system but negative work instead. The net force acting of the system is zero since the upward tension in the string suspending the object is equal to the weight of the object but acting in the opposite direction. As a result they cancel out. In the equation above the effect of the tension force on the object has been neglected or not taken into consideration. For the mechanical energy E to be conserved, the work done by this tension force must be included into the equation. Otherwise it would seem as though energy has been generated in some manner that is equal in magnitude to the work done by the tension force.

The conserved form of the equation is given by

E = K + Ug + Wother.

In this case Wother = work done by the tension force.

In that form the total mechanical energy is conserved.

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )