Question

prove that the mechanical energy (KE+PE) of a freely falling body always remain constant​

Answer 1

Answer:

The body has moved a distance x from A. If the body reaches the position C. 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.

Hence Proved

Answer 2

Mechanical energy of free falling body remains constant is proved below -

Let a body of mass m be at height A. It falls freely and at height B, which is (X) metre from a and (h-x) metre from c, the velocity of falling is v. C is the point at Earth's surface and the velocity here is v'.

At point A, Kinetic energy of body is zero, hence it only has potential energy.

E = m*g*h.

At point B, body will have both kinetic energy and potential energy.

E = 0.5*m*v² + mg(h-x)

Calculating v² -> v² = u² +2*g*X. Since u is zero, v² = 2*g*X.

Keeping the value in equation

E = 0.5*m*2*g*X + m*g*h - m*g*X

E = m*g*h.

At point C, body will have only kinetic energy, since height is zero.

E = 0.5*m*v'²

Again calculating v'² -> v'² = u² +2*g*h. Since u is zero, v'² = 2*g*h.

Keeping the value in equation

E = 0.5*m*2*g*h

E = m*g*h.

Thus, mechanical energy at all the three points namely A, B and C is mgh, which remains conserved.