Mechanical energy remains constant throughout the fall, what will be the speed when height is h/4
Answers
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.
At A,
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.
At B,
velocity v2 = u2 + 2as
applying, v2 = 0 + 2ax = 2ax
Kinetic energy Ek = 1/2 mv2 = 1/2 m x 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.
At C,
Potential energy Ep = 0
Velocity of the body C is
v2 = u2 + 2as
u = 0, a = g, s = h
applying v2 = 0 + 2gh = 2gh
kinetic energy Ek =1/2 mv2=1/2 m x 2gh= mgh
Total energy at C
E = Ep + Ek
E = 0 + mgh
E = mgh
Thus we have seen that sum of potential and kinetic energy of freely falling body at all points remains same. Under the force of gravity, the mechanical energy of a body remains constant.
K + U = cons.
1/2mv2 = mgH
V = root (2gH)
= root (gh/2) [ H =h/4]