# Why is potential energy increasing?

This is the definition of Work that I understand.

Work: Work is any mechanical process that involves transfer of energy from one structure to another.

My question is, when we are lifting a block of mass $m$ slowly, the $\Delta K = 0$ implying total work = 0. There is no net work done on the block but its potential energy is increasing, why?

• It's not clear to me why your definition means that you aren't doing work by lifting it up. No matter how you lift the block; you will be taking energy from somewhere to do it. Your example showing kinetic energy is constant doesn't really change that. – JMac Mar 15 '18 at 0:06

You have stumbled upon a very important point of mechanics. It is essential to define your system, and identify what forces are internal, and which are external.

If you define your system to be the block and the earth then the force of your hand is an external force, and the force of gravity is internal. External forces can do work on the system, and raise its energy.$$\Delta E = W_\mathrm{external}$$ Internal forces can exchange energy among sub-systems. Internal (conservative) forces are also responsible for the existence of potential energy. Potential energy is defined as the negative of internal work done by a conservative force$$\Delta U = -W_\mathrm{internal}$$ In order for there to be internal work, the system must comprise at least two objects: one to be the object of a force, the other to be the agent. Two interacting objects are necessary for potential energy to exist. Unfortunately, many introductory expositions of potential energy fail to present the correct definition, and these points are not discussed. So if the system is the block and the Earth, your hand does external work, which raises the energy of the system.$$\Delta E = W_\mathrm{external}=F_\mathrm{hand}h$$

There is no change in kinetic energy, as you point out. The only internal energy is the potential energy gained by the internal work done by gravity$$\Delta U = -W_\mathrm{internal} = -(-F_\mathrm{gravity}h)$$ But by Newton's Second and Third laws the magnitude of the force of the hand is equal to the magnitude of the force of the Earth, and$$\Delta U = mgh$$

On the other hand, if you define your system as the block, then both the force of gravity and the force of your hand are external. There is no other object in the system that can supply internal work. Potential energy is not defined. Your hand does positive work, and gravity does the same amount of negative work. The net work is zero. The kinetic energy does not change, and potential energy does not exist so that it is not even something that can be considered.

To solve a mechanics problem, it is essential to carefully define your system. How you decide what is inside the system and what is outside doesn't matter, you'll get the same answer regardless of where you define the boundary between system and environment. After all, nature doesn't care about an imaginary boundary. However, the energy accounting will be different.

Potential energy is increasing because you are bringing the block of mass higher into the air. The higher it is from the ground, the more potential energy it has.

As you point out "There is no net work done on the block". The work you do, which is positive, is the same work gravity does but in the negative side. The Work Energy Theorem Holds Here, as it by definition accounts for all conservative and non-conservative forces.

Going further, we call this negative work by gravity as the potential energy stored in the block.

Right$\,$! $\;$ Work is not defined the way you state. The definition is Force times Distance$\,$: $W = F \times s$.$\,$ So if you lift a block of mass 10 kg up to a point 10 m higher, then the force $F = mg = 10 \, {\rm kg} \cdot 9.81\, {\rm m/s^2}$,$\,$ and the distance is $\; s = 10 {\rm m}\, . \,$ The work done then equals $W= 981 \, {\rm Joule}$.