Question

State Law of Conservation of Energy and express it in the form of an equation for a body of

mass m falling from a point A at height h, above the ground at (a) A, (b) B at a height from

ground, (c) C.​

Answer 1

hope above answer may be help u

Answer 2

Answer:

⭐❤☄Energy may change in form or be transferred from one system to another, but the total remains the same. When all forms of energy are considered, conservation of energy is written in equation form as KEi + PEi + Wnc + OEi = KEf + PEf + OEf, where OE is all other forms of energy besides mechanical energy.☄❤⭐

Let a body of mass m be droped from a point A at a height H. P.E. at A = mgh K.E. = 0 Total energy at A = mgh As it reaches B, it would have lost some P.E. and gained K.E. Velocity on reaching B =  √ 2 g h 2gh P.E. at B = mg(H – x) K.E. =  1 2 m v 2 B 12mvB2 = mgx Total energy at B = mg(H – x) + mgx = mgH On reaching the ground C the mass must have gained a velocity √ 2 g H 2gH and the P.E. must be zero. P.E. at C = 0 K.E. at C =  1 2 m v 2 C 12mvC2 =  1 2 m ( 2 g H ) 12m(2gH) = mgH Total energy at C = mgH Thus it is proved that the total energy at any point in its path is mgHRead more on Sarthaks.com - https://www.sarthaks.com/1034060/a-body-of-mass-m-is-released-in-vacuum-from-the-position-a-at-a-height-h-above-the-ground