Question

When an object is dropped from a height, and then it reaches the ground, then potential energy is not equal to kinetic energy in each case, i.e., PE ≠ KE....(1)
But, KE = ½mv² = ½ × m × 2gh [since v² – u² = 2gh => v² = 2gh, when u = 0].
=> KE = mgh = PE....(2)
So, equation (1) contradicts equation (2).
Which is correct then?
Is KE = PE, or is KE ≠ PE?​

Answer 1

Answer:

Equation 2 is incorrect

Explanation:

By KE = ½mv² = ½ × m × 2gh, h represents the height that the object has fallen from its original height.

But by PE = mgh, h represents the height of the object from the ground.

Since h represents different quantities, KE ≠ PE

Answer 2

Answer:

KE can be equal to PEin some cases at some instant but it is purely coincidental. You are just applying the formula. This is not the way the concept of work and energy works. You have to apply the conservation of mechanical energy at two instants. First, when you are dropping an object and Second, when the object reaches the ground. Read full explanation below to understand this.

Explanation:

You need to first understand the concept of work and energy.

Let's start,

First is Kinetic Energy. KE can be defined as an energy which is possessed by a body by virtue of it's mototion. It can be represented mathematically as

$kinetic \: energy = \frac{1}{2} mv {}^{2}$

here m is the mass of the object and v is the velocity at that instant when you want to calculate the value of KE.

Second one is Potential Energy. There can be many types of potential energies. The one which you need for this problem is Gravitational Potential Energy and it can represented mathematically as

$potential \: energy = mgh$

here m is the mass of the object and h is the height measured from a reference level which, in this case, is ground.

Mechanical Energy is a term which is equal to PE+KE at any instant.

Now Principle of conservation of mechanical energy says that if a system is governed by only conservative forces, it's ME (KE+PE) remains constant.

here only mg force is acting on the object which is a conservative force. So ME is conserved throughout the motion.

hence you can apply ME conservation between any two instants during the motion.

Now you have to find the KE+PE at that instant when the object is dropped and then equate it with the value of KE+PE at that instant when object just reaches the ground.

Initially, when object is dropped

$kinetic \: energy = \frac{1}{2} mu {}^{2} = 0$

as u=0

$potential \: energy = mgh$

h is the height from where you are dropping the object.

Hence, Initially ME is

$mechanical \: energy = kinetic \: energy + potential \: energy$

$mechanical \: energy = 0 + mgh$

$mechanical \: energy = mgh$

Finally, when the object reaches ground

$kinetic \: energy = \frac{1}{2} mv {}^{2}$

v is the final velocity here

$potential \: energy = mgh = 0$

as h is zero now

Hence, Finally ME is

$mechanical \: energy = \frac{1}{2} mv {}^{2} + 0$

$mechanical \: energy = 0$

Now according to the principle of conservation of ME, Intial ME = Final ME

so,

$initial \: mechanical \: energy = final \: mechanical \: energy$

$mgh = \frac{1}{2} mv {}^{2}$

Here mgh is Initial PE and 1/2mv^(2) is final KE.

From above equation you get

$v = \sqrt{2gh}$

Conclusion:

PE and KE are changing every moment so there can be a moment when PE and KE are equal but it is not the case with every instant.

If you put v=√2gh in the formula of Kinetic energy, you get KE at that instant when the velocity was √2gh i.e when the object was about to touch the ground and you get it to be equal to mgh which is PE at that instant when you dropped the object i.e initial PE

So for an instant you cannot say that KE = PE