Show that the sum of kinetic energy and potential energy(i.e., total mechanical energy) is always conserved in the case of freely falling body under gravity (with air resistance neglected) from a height h by finding it when (i) the body is at the top, (ii) the body has fallen a distance x, (iii) the body has reached the ground.
Show that the sum of kinetic energy and potential energy(i.e., total mechanical energy) is always conserved in the case of freely falling body under gravity (with air resistance neglected) from a height h by finding it when (i) the body is at the top, (ii) the body has fallen a distance x, (iii) the body has reached the ground.
It may be shown that in the absence of external frictional force the total mechanical energy of a body remains constant.
Let a body of mass m falls from a point A, which is at a height h from the ground as shown in fig.
1)At top,
Kinetic energy kE = 0
Potential energy Ep = mgh
Total energy E = Ep + Ek = mgh + 0= mgh
During the fall, the body is at a position B. The body has moved a distance x from A.
2)At has fallen a distance x
velocity 
+ 2as
applying, 
= 0 + 2ax = 2ax
Kinetic energy Ek =
= 
m × 2gx = mgx
Potential energy Ep = mg (h – x)
Total energy E = Ep + Ek = mg (h-x) + mgx = mgh – mgx + mgx= mgh
If the body reaches the position C.
3)At ground,
Potential energy Ep = 0
Velocity of the body C is
+ 2as
u = 0, a = g, s = h
applying v2 = 0 + 2gh = 2gh
kinetic energy Ek =
m × 2gh = mgh
Total energy at ground
E = Ep + Ek
E = 0 + mgh
E = mgh
Thus we have seen that the sum of potential and kinetic energy of the freely falling body at all points remains the same. Under the force of gravity, the mechanical energy of a body remains constant.
It may be shown that in the absence of external frictional force the total mechanical energy of a body remains constant.
Let a body of mass m falls from a point A, which is at a height h from the ground as shown in fig.
1)At top,
Kinetic energy kE = 0
Potential energy Ep = mgh
Total energy E = Ep + Ek = mgh + 0= mgh
During the fall, the body is at a position B. The body has moved a distance x from A.
2)At has fallen a distance x
velocity + 2as
applying, = 0 + 2ax = 2ax
Kinetic energy Ek = = m × 2gx = mgx
Potential energy Ep = mg (h – x)
Total energy E = Ep + Ek = mg (h-x) + mgx = mgh – mgx + mgx= mgh
If the body reaches the position C.
3)At ground,
Potential energy Ep = 0
Velocity of the body C is
+ 2as
u = 0, a = g, s = h
applying v2 = 0 + 2gh = 2gh
kinetic energy Ek = m × 2gh = mgh
Total energy at ground
E = Ep + Ek
E = 0 + mgh
E = mgh
Thus we have seen that the sum of potential and kinetic energy of the freely falling body at all points remains the same. Under the force of gravity, the mechanical energy of a body remains constant.
