Under what circumstances the work done by an external force gets stored as potential energy?

Question

Consider earth and a block as a system. The block is accelerating downward under gravity. Now I apply a force against gravity which is less than the force of gravity. Would the work done against gravity by the external force still be stored as potential energy? If yes, I think we can say that total mechanical energy is conserved. But when an external force does work mechanical energy is not conserved.

Under what circumstances does the work done against internal conservative force gets stored as potential energy?

Why did we define the notion of potential energy when we could easily have explained the situation with work-energy theorem?

Please correct me if I'm wrong.

Answer 1

You have clearly defined the system - Earth & block.

The gravitational forces, block on Earth and Earth on block are internal forces.
The force applied against the gravitational force is an external force.

You can then apply energy conservation in a number of ways.
For example, one way is to evaluate the changes in kinetic energy and the work done by all (both external and internal) forces and not mention potential energy.
Another way is to evaluate the changes in kinetic energy, the work done by the external force and the change in the gravitational potential energy of the system.

Making the assumption that the mass of the Earth is much greater than the mass of the block and so the distance moved by the block, $\Delta h$, is much greater than the distance moved by the Earth, the work done by the internal forces $mg \times \Delta h$ comes out to be the same as the change in the gravitational potential energy of the system.

You have described a relatively simple system.
If the force is conservative then force and the rate of change of potential energy are related but it is often the case that using the concept of potential energy makes things easier to understand and evaluate particularly as force is a vector and potential energy is a scalar.
If there is a spring involved you can choose between using $F=-ke$ and evaluating the work done or using the potential energy approach with $U_{\rm elastic} = \frac 12 k e^2$.

Under what circumstances does the [external] work done against internal conservative force gets stored as potential energy?
If you mean that the potential energy increases, then when the external work done on a system is positive, ie the block moves upwards in your example.
In the case of your falling block, with or without the external force for a given change in height of the block the change is gravitational potential energy is the same, all that differs is the change in the kinetic energy of the block.

Answer 2

I think we can say that total mechanical energy is conserved

This is incorrect - the mechanical energy of the Earth/block system changes.

Let's assume that the block starts from rest; the block has mass $m$ which is very much smaller than the mass of the Earth (so we can treat the Earth as being stationary throughout the motion); the direction of the force $\vec F$ is vertically upwards; and $|\vec F| < mg$.

Then the block accelerates downwards with acceleration

$\displaystyle a = g - \frac {|\vec F|} m$

After a time $t$ the block has velocity $v = at$ and it has fallen a distance $d = \frac 1 2 a t^2$. It therefore has kinetic energy $\frac 1 2 mv^2 = \frac 1 2 ma^2t^2 = mad$ and the Earth/block system has lost potential energy $mgd$. The total change in mechanical energy of the Earth/block system is therefore

$\Delta KE_{Earth} + \Delta KE_{block} + \Delta PE = 0 + mad - mgd = (a-g)md$

But we know that $a-g = - \frac {|\vec F|} m$, so

$\Delta KE + \Delta PE = - |\vec F|d$

So the mechanical energy of the Earth/block system is not conserved - it has changed by an amount $- |\vec F|d$ which is the work done by the force $\vec F$; note that the sign is negative because $d$ is in the opposite direction to the direction of $\vec F$.

Answer 3

While we can explain the gain in kinetic energy in terms of work = force $\times$ distance, the potential energy interpretation is more general and can be applied where for example the potential energy is chemical and the work energy relationship is not so obvious.

Here is a numeric example to show energy is conserved when taking potential energy into account:

I am using a system where measurements are made relative to an observer at rest with the pericentre of the system and both the object and the Earth are moving, rather than make the approximation that the Earth is not moving.

Quantities relating to the Earth have a E subscript and quantities relating to the block have a B subscript.

Assuming $m_E = 5.97E24\ kg\ $, $m_B = 5 kg\ $, $a_B = 9.807 \ m/s^2\ $, $R= 6.371009 E6\ m \ $, $t = 3 s $

$v_B = a_0 t = 29.421\ m/s$.

Assuming conservation of momentum is valid, then:

$ m_E \ v_E = m_B\ v_B $

where $v_E ,\ v_B $ are the final velocities.

$v_E = v_B \frac{m_B}{m_E} = 14.92814/E24\ m/s$

$a_E = v_E / t = a_B \frac{m_B}{m_E} = 1.64271/E24\ m/s^2$

$d_B = 1/2 \ a_B t^2 = 44.1315 \ m$

$d_E = 1/2 \ \_E t^2 = 7.39221 \ m$

$KE_B = 1/2*m_B*v_B^2 = 432.798\ J$

$PE_B = m_B\ a_B\ d_B = 432.798\ J$

$KE_E = 1/2\ m_E*v_E^2 = 72.4954/E24\ J$

$PE_E = m_E\ a_E\ d_E = 72.4954/E24\ J$

So $\Delta KE_B - \Delta PE_B = 0$

and $\Delta KE_E - \Delta PE_E = 0$

so it would appear energy is conserved.