Question

Show that the sum of kinetic energy and potential energy that is total mechanical energy is always concern in the case of a freely falling body on the gravity with a resistance neglected from a height h by finding it when number 1 the body is at the top to the body has worked on a distance x 3 the body has reached the ground

Answer 1

Explanation:

Let a body of mass m be falling freely under gravity from a height h above the ground (i.e., from position A). Let us now calculate the sum of kinetic energy K and potential energy U at various positions, say at A (at height h above the ground), at B (when it has fallen through a distance x) and at C (on the ground).

(i) At the position A (at height h above the ground)

Initial velocity of body= 0 (since body is at rest at A)

Hence, kinetic energy K =0

Potential energy U = mgh

Hence total energy = K + U = 0 + mgh = mgh .. -----(i)

(ii) At the position B (when it has fallen a distance x): Let v1 be the velocity acquired by the body at B after falling through a distance x. Then u = 0, S = x, a = g

From equation v2 = u2 + 2As V12 = 0 + 2gx = 2gx Hence, Kinetic energy K = 1/2 mv 2/1 Now at B, height of body above the ground = h – x Hence, potential energy U = mg (h – x) Hence total energy = K + U = mgx + mg (h – x) = mgh ----------- (ii)

(iii) At the position C (on the ground): Let the velocity acquired by the body on reaching the ground be v. Then u = 0, S = h, a = g From equation: v2 = u2+2aS v2 = 02 + 2gh v2 = 2gh Hence, kinetic energy K = 1/2 mv2 = 1/2 m (2gh) = mgh And potentiall energy U = 0 (a the ground when h = 0) Hence total energy = K+U = mgh + 0 = mgh ------ (iii)

Thus from equation (i), (ii) and (iii), we note that the total mechanical energy i.e., the sum of kinetic energy and potential energy always remain constant at each point of motion and is equal to initial potential energy at height h.