Consider the case of a freely falling body given in the following figures:
At A
Kinetic energy=0
Potential energy=mgh
At B
Kinetic energy=mgx
At C
Kinetic energy=mgh
Potential energy=0
a) Find the potential energy of the body at B.
b) Find the total energy at A,B and C.
c) Is there any variation in total energy? What do you infer from the result?

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Solution

(a) Since total mechanical energy is conserved at each point in path,
$K E_{B} + P E_{B} = K E_{A} + P E_{A}$
$m g x + P E_{B} = 0 + m g h$
$P E_{B} = m g (h - x)$
(b)Total energy at $A = K E_{A} + P E_{A} = m g h$
Total energy at $B = K E_{B} + P E_{B} = m g h$
Total energy at $C = K E_{C} + P E_{C} = m g h$
(c) There is no variation in total energy.This shows that the total energy is always conserved.

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(a) Since total mechanical energy is conserved at each point in path,KEB+PEB=KEA+PEAmgx+PEB=0+mghPEB=mg(h−x)(b)Total energy at A=KEA+PEA=mghTotal energy at B=KEB+PEB=mghTotal energy at C=KEC+PEC=mgh(c) There is no variation in total energy.This shows that the total energy is always conserved.