Mathematically prove that the total mechanical energy of a freely falling body is constant at all positions .
Asked by Taruna
| 10th Sep, 2015,
06:44: PM

Let us consider an object of mass, ‘m’ is thrown upwards to a height ‘h’ above the ground level.
At point A:
The body must be at rest and v = 0
PEA = mgh and KEA = 0
Total energy = PEA + KEA = mgh ......... (Equation 1)
When the body drops down, let us say it covers a distance ‘x’ at point C. Thus, its height from the ground is ‘h – x’.
At point C:
PEC = mg(h-x) = mgh – mgx ........ (Equation 2)
At point B:
The body hits the ground u = 0 and h = 0
So, it possesses only kinetic energy.
PEA = 0

Thus, from equation (1), (5) and (7) that the total energy remains constant.
Let us consider an object of mass, ‘m’ is thrown upwards to a height ‘h’ above the ground level.
At point A:
The body must be at rest and v = 0
PEA = mgh and KEA = 0
Total energy = PEA + KEA = mgh ......... (Equation 1)
When the body drops down, let us say it covers a distance ‘x’ at point C. Thus, its height from the ground is ‘h – x’.
At point C:
PEC = mg(h-x) = mgh – mgx ........ (Equation 2)

The body hits the ground u = 0 and h = 0
So, it possesses only kinetic energy.
PEA = 0
Thus, from equation (1), (5) and (7) that the total energy remains constant.
Answered by Yashvanti Jain
| 11th Sep, 2015,
10:25: AM
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