Confirmation on Potential energy of an object

Question

I very well understand that a lot of people have asked the question "where does potential energy come from" on this site, but mine is more of clarification.

If an object was propelled from the ground to a certain height h, it's potential energy at that height (h) is said to be mgh. Since the final K.E and initial K.E is 0, total work done is also 0. Hence why is P.E = mgh and not 0. I have a possible explanation, but I'm not exactly sure. The force required to propel an object on the ground, to a height (h) , has to be greater than it's weight. Let's label this extra force f', therefore;

F= mg+f'. Where F is the propelling force.

Hence the work done by this force is;

W= mgh+f'h

According to the work-energy theorem,

Kf-Ki= Net work done

Net work done = Net force * distance

Net force = weight+ propelling force

.'. Net force= -mg+mg+f'

Hence: Net force= f'

.'. Net work done= f'h

Therefore Kf-Ki= f'h

Since Kf and Ki = 0,

f'h= 0

Recall, Work done= mgh + f'h

Since f'h=0,

Work done = mgh.

Is any of this correct? If not, Where did I go wrong?

Answer 1

Suppose the object starts out at height $0$ with an initial vertical velocity $u$. So it's kinetic energy at height $0$, which we will denote by $KE(0)$ is $\frac 1 2 mu^2$. We don't know its potential energy at height $0$, but we will just call this $PE(0)$.

We know that the object is decelerating at a rate of $-g$ due to gravity. If its velocity at height $h$ is $v(h)$ then its velocity at height $h$ is given by

$v^2(h) = v^2(0) - 2gh = u^2 - 2gh$

so its kinetic energy at height $h$ is

$KE(h) = \frac 1 2 m v^2(h) = \frac 1 2 m(u^2 - 2gh) = \frac 1 2 mu^2 - mgh = KE(0) - mgh$

From conservation of energy we know that

$PE(h) + KE(h) = PE(0) + KE(0) \\ \Rightarrow PE(h) = PE(0) + KE(0) - KE(h) = PE(0) + \frac 1 2 mu^2 - (\frac 1 2 mu^2 - mgh) = PE(0) + mgh$

We still don't know $PE(0)$, but we do know that the difference in potential energy between height $0$ and height $h$ is $\Delta PE = PE(h) - PE(0) = mgh$. So, no matter what value we assign to $PE(0)$, we know that the additional potential energy that the object has at height $h$ is $mgh$.

Alternatively, you can think of the change in potential energy as being the negative of the work done by gravity on the object between height $0$ and height $h$. We sometimes call this the work done against gravity.

Answer 2

First, here potential energy means gravitational potential energy. It is numerically equal to the work done by the(conservative) gravitational field, and not equal to the change in kinetic energies( except when the object is in free fall).The work energy theorem implies that the work done by all the forces acting on the body is equal to the change in kinetic energy of the body. Regarding the calculation, you have assumed the force to be constant, but if the force is constant, the object would continue moving without coming to rest. You must assume a variable force such that work done by the force is 0 for the equations to be valid. Also its not that the initial and final kinetic energies have to be zero, its their change that has to be zero. If you lift an object from the ground up, gravitational force would do negative work on it, and hence the object would gain GPE.

Answer 3

If an object was propelled from the ground to a certain height h, it's potential energy at that height (h) is said to be mgh. Since the final K.E and initial K.E is 0, total work done is also 0

If you assume that at height $h$, it doesn't have any kinetic energy, you can say initial kinetic energy is transferred to potential energy. Here what you should be careful is that when you throw an object with some velocity $v$, it doesn't have zero initial kinetic energy. Therefore, the potential energy is not zero at height $h$.