Show that the sum of kinetic energy and potential energy (i.e., total mechanical energy) is always conserved in the case of a 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.

Question

Show that the sum of kinetic energy and potential energy (i.e., total mechanical energy) is always conserved in the case of a 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.

KnowledgeBoat · Accepted Answer

Show that kinetic energy and potential energy sum is always conserved for freely falling body under gravity. Work, Energy, Power Concise Physics Class 10 Solutions.

We know that, Kinetic energy + potential energy = constant So, when a body falls from a height h under free fall (i) At the position A height h Initial velocity = 0 Kinetic energy K = 0 Potential energy U = mgh As, Total energy = KE + PE Total energy = 0 + mgh Total energy = mgh (1) (ii) At the position B when it has fallen a distance x. Then, velocity at B = v₁ Then u = 0, s = x, a = g From equation of motion: v² = u² + 2aS v₁² = 0 + 2gx = 2gx ∴ Potential energy U = mg (h – x) Hence, total energy = K + U = mgx + mg (h – x) = mgh Total energy = mgh (2) (iii) At position C (on the ground) Let the velocity acquired by the body on reaching the ground be v₂. Then u = 0, s = h, a = g We know, v²= u² + 2aS v₂² = 0 + 2gh v₂² = 2gh And potential energy U = 0 (at the ground when h = 0) So, total energy = K + U = mgh + 0 Total energy = mgh (3) Thus, from equation (1), (2) and (3) 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 it is equal to the initial potential energy at height h.

Physics

Work, Energy & Power

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